JEE Practice Questions (NAT)

Hint

(d)

\(
\begin{aligned}
f(\theta) & =\frac{1-\sin 2 \theta+\cos 2 \theta}{2 \cos 2 \theta} \\
& =\frac{(\cos \theta-\sin \theta)^2+\left(\cos ^2 \theta-\sin ^2 \theta\right)}{2(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)} \\
& =\frac{\cos \theta}{\cos \theta+\sin \theta} \\
& =\frac{1}{1+\tan \theta} \\
f\left(11^{\circ}\right) \cdot f\left(34^{\circ}\right) & =\frac{1}{\left(1+\tan 11^{\circ}\right)} \times \frac{1}{\left(1+\tan 34^{\circ}\right)} \\
& =\frac{1}{\left(1+\tan 11^{\circ}\right)} \times \frac{1}{\left(1+\tan \left(45^{\circ}-11^{\circ}\right)\right)} \\
& =\frac{1}{\left(1+\tan 11^{\circ}\right)} \times \frac{1}{\left(1+\frac{1-\tan 11^{\circ}}{1+\tan 11^{\circ}}\right)} \\
& =\frac{1}{\left(1+\tan 11^{\circ}\right)} \times \frac{\left(1+\tan 11^{\circ}\right)}{2}=\frac{1}{2}
\end{aligned}
\)

\(
8 f\left(11^{\circ}\right) \cdot f\left(34^{\circ}\right)=4
\)

Hint

(c)
\(
\begin{aligned}
& f(x)=2(7 \cos x+24 \sin x)(7 \sin x-24 \cos x) \\
& r \cos \theta=7 ; r \sin \theta=24 \\
& r^2=625 ; \tan \theta=\frac{24}{7}
\end{aligned}
\)
\(
\begin{aligned}
f(x) & =2 r \cos (x-\theta) \times r \sin (x-\theta) \\
& =r^2(\sin 2(x-\theta)) \\
f(x)_{\max } & =25^2 \Rightarrow(f(x))^{1 / 4}=5
\end{aligned}
\)

Hint

(a) 

\(
\begin{aligned}
& \tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)=-\frac{b}{a} ; \tan \left(\frac{A}{2}\right) \times \tan \left(\frac{B}{2}\right)=\frac{c}{a} \\
& A+B=90^{\circ} \Rightarrow \frac{A+B}{2}=45^{\circ} \\
& \Rightarrow \tan \left(\frac{A+B}{2}\right)=1=\frac{-\frac{b}{a}}{1-\frac{c}{a}} \\
& \Rightarrow 1-\frac{c}{a}=-\frac{b}{a} \\
& \Rightarrow a+b=c \\
& \Rightarrow \frac{a+b}{c}=1
\end{aligned}
\)

Hint

(d)
\(
\begin{aligned}
(1+\tan \theta)\left[1+\tan \left(45^{\circ}-\theta\right)\right]= & (1+\tan \theta)\left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) \\
& =(1+\tan \theta)\left(\frac{2}{1+\tan \theta}\right)=2
\end{aligned}
\)
\(
\begin{aligned}
&\text { Hence, L.H.S. is equal to }\\
&\begin{aligned}
& 2\left(1+\tan 5^{\circ}\right)\left(1+\tan 40^{\circ}\right)\left(1+\tan 10^{\circ}\right)\left(1+\tan 35^{\circ}\right)\left(1+\tan 15^{\circ}\right)\left(1+\tan 30^{\circ}\right)\left(1+\tan 20^{\circ}\right)\left(1+\tan 25^{\circ}\right) \\
& =2 \times 2^4=2^5
\end{aligned}
\end{aligned}
\)
\(
2^k=2^5; k =5
\)

Hint

(b)

\(
\begin{aligned}
\sqrt{3}\left|\frac{\frac{-2 \sin \left(40^{\circ}\right) \cos \left(40^{\circ}\right)}{\cos \left(80^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)}}{\frac{\cot \left(20^{\circ}\right)+\tan \left(80^{\circ}\right)}{\cot \left(20^{\circ}\right)}}\right| & =\sqrt{3}\left|\frac{\tan \left(20^{\circ}\right)-\tan \left(80^{\circ}\right)}{1+\tan 20^{\circ} \tan 80^{\circ}}\right| \\
& =\sqrt{3} \tan \left(60^{\circ}\right)=3
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \text { Let } x+5=14 \cos \theta \text { and } y-12=14 \sin \theta \\
& \begin{aligned}
\therefore x^2+y^2 & =(14 \cos \theta-5)^2+(14 \sin \theta+12)^2 \\
& =196+25+144+28(12 \sin \theta-5 \cos \theta) \\
& =365+28(12 \sin \theta-5 \cos \theta)
\end{aligned} \\
& \left.\therefore \sqrt{x^2+y^2}\right|_{\min }=\sqrt{365-28 \times 13}=\sqrt{365-364}=1
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& \cot x+\cot y=49 \\
& \Rightarrow \frac{1}{\tan x}+\frac{1}{\tan y}=49 \\
& \Rightarrow \frac{\tan y+\tan x}{\tan x \tan y}=49 \\
& \Rightarrow \tan x \tan y=\frac{\tan x+\tan y}{49}=\frac{42}{49}=\frac{6}{7} \\
& \Rightarrow \tan (x+y)=\frac{42}{1-(6 / 7)}=\frac{42}{1 / 7}=294 \text { which is divisible by } 2,3 \text { and } 7 \text { but not by } 5 .
\end{aligned}
\)

Hint

(b)

From the given equations, we have
\(
(2 \cos a+9 \cos d)^2=(6 \cos b+7 \cos c)^2
\)
And \((2 \sin a-9 \sin d)^2=(6 \sin b-7 \sin c)^2\)
Adding; we have \(36 \cos (a+d)=84 \cos (b+c)\)
\(
\Rightarrow \frac{\cos (a+d)}{\cos (b+c)}=\frac{7}{3}
\)
\(
3 \frac{\cos (a+d)}{\cos (b+c)}=7
\)

Hint

(d)

\(
\begin{aligned}
&\text { Since } \cos A+\cos B=0\\
&\begin{aligned}
& \Rightarrow A+B=\pi \\
& \therefore B=\pi-A \\
& \Rightarrow \sin A+\sin (\pi-A)=1 \\
& \Rightarrow \sin A=\frac{1}{2} \\
& \Rightarrow A=30^{\circ} \text { and } B=150^{\circ} \text { or } A=150^{\circ} \text { and } B=30^{\circ} \\
& \Rightarrow 12 \cos 60^{\circ}+4 \cos 300^{\circ}=8
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
& 5 \frac{2 \tan \beta}{1+\tan ^2 \beta}=3 \frac{2 \tan \alpha}{1+\tan ^2 \alpha} \\
& \Rightarrow \frac{5 \tan \beta}{1+\tan ^2 \beta}=\frac{3 \tan \alpha}{1+\tan ^2 \alpha} \dots(i)
\end{aligned}\\
&\text { Substituting } \tan \beta=3 \tan \alpha \text {, we have }\\
&\begin{aligned}
& \frac{5 \times 3 \tan \alpha}{1+9 \tan ^2 \alpha}=\frac{3 \tan \alpha}{1+\tan ^2 \alpha} \\
& \Rightarrow 5+5 \tan ^2 \alpha=1+9 \tan ^2 \alpha \\
& \Rightarrow 4 \tan ^2 \alpha=4 \\
& \Rightarrow \tan \alpha=1, \text { i.e., } \tan \beta=3 \\
& \therefore \tan \alpha+\tan \beta=4
\end{aligned}
\end{aligned}
\)

Hint

(b)
\(
\begin{aligned}
&\text { Let } \theta=\frac{\pi}{16} \Rightarrow 8 \theta=\frac{\pi}{2}\\
&\begin{aligned}
y= & \tan \theta+\tan 5 \theta+\tan 9 \theta+\tan 13 \theta \\
\therefore y & =(\tan \theta-\cot \theta)+(\tan 5 \theta-\cot 5 \theta) \\
& \quad[\operatorname{astan} 13 \theta=\tan (8 \theta+5 \theta)=-\cot 5 \theta \text { and } \tan 9 \theta=\tan (8 \theta+\theta)=-\cot \theta] \\
& =(\tan \theta-\cot \theta)+(\cot 3 \theta-\tan 3 \theta) \\
& =\frac{\sin ^2 \theta-\cos ^2 \theta}{\sin \theta \cos \theta}+\frac{\cos ^2 3 \theta-\sin ^2 3 \theta}{\sin 3 \theta \cos 3 \theta}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow y & =2\left[\frac{\cos 6 \theta}{\sin 6 \theta}-\frac{\cos 2 \theta}{\sin 2 \theta}\right] \\
& =2\left[\frac{\sin 2 \theta \cos 6 \theta-\cos 2 \theta \sin 6 \theta}{\sin 6 \theta \sin 2 \theta}\right] \\
& =-2\left[\frac{\sin 4 \theta}{\cos 2 \theta \sin 2 \theta}\right]=-4 \quad\left(\because 6 \theta=\frac{\pi}{2}-2 \theta\right)
\end{aligned}
\)
\(
\text { Hence, absolute value }=4 \text {. }
\)

Hint

(a)
\(
\begin{aligned}
& \cos 290^{\circ}=\sin 20^{\circ} ; \sin 250^{\circ}=-\sin 70^{\circ}=-\cos 20^{\circ} \\
& \Rightarrow \frac{1}{\sin 20^{\circ}}-\frac{1}{\sqrt{3} \cos 20^{\circ}}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& =\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sqrt{3} \sin 20^{\circ} \cos 20^{\circ}} \\
& =\frac{2\left[\sin 60^{\circ} \cos 20^{\circ}-\sin 20^{\circ} \cos 60^{\circ}\right]}{\sqrt{3} \sin 20^{\circ} \cos 20^{\circ}} \\
& =\frac{4 \sin 40^{\circ}}{\sqrt{3} \sin 40^{\circ}}=\frac{4 \sqrt{3}}{3}
\end{aligned}\\
&\text { Hence, the greatest integer less than or equal to is } 2
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \sin ^6 x+\cos ^6 x=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x-\sin ^2 x \cos ^2 x\right) \\
& =1-3 \sin ^2 x \cos ^2 x=1-\frac{3(\sin 2 x)^2}{4} \\
& \Rightarrow y=\frac{4}{4-3(\sin 2 x)^2} \\
& \Rightarrow y_{\max }=\frac{4}{4-3(1)}=4
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
\cos ^2\left(45^{\circ}+x\right)+(\sin x-\cos x)^2 & =\left[\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}}\right]^2+(\sin x-\cos x)^2 \\
& =\frac{3}{2}(1-\sin 2 x)=\frac{3}{2}(1-(-1))
\end{aligned}\\
&\text { Hence, the maximum value is }=\frac{3}{2}(1-(-1))=3
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& \text { Nr. }=\left(\sin ^2 t+\cos ^2 t\right)^2-2 \sin ^2 t \cos ^2 t-1=-2 \sin ^2 t \cos ^2 t \\
& \text { Dr. }=\left(\sin ^2 t+\cos ^2 t\right)^3-3 \sin ^2 t \cos ^2 t-1=-3 \sin ^2 t \cos ^2 t
\end{aligned}
\)
\(
9 \frac{\sin ^4 t+\cos ^4 t-1}{\sin ^6 t+\cos ^6 t-1}=6
\)

Hint

(c)

\(
\begin{aligned}
\frac{1}{\sin 10^{\circ}}+\frac{1}{\sin 50^{\circ}}-\frac{1}{\sin 70^{\circ}} & =\frac{1}{\cos 80^{\circ}}+\frac{1}{\cos 40^{\circ}}-\frac{1}{\cos 20^{\circ}} \\
& =\frac{\cos 40^{\circ} \cos 20^{\circ}+\cos 80^{\circ} \cos 20^{\circ}-\cos 40^{\circ} \cos 80^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}} \\
& =8\left[\cos 20^{\circ}\left(\cos 40^{\circ}+\cos 80^{\circ}\right)-\cos 40^{\circ} \cos 80^{\circ}\right] \\
& =8\left[2 \cos 20^{\circ} \cos 60^{\circ} \cos 20^{\circ}-\cos 40^{\circ} \cos 80^{\circ}\right] \\
& =4\left[2 \cos ^2 20^{\circ}-2 \cos 40^{\circ} \cos 80^{\circ}\right] \\
& =4\left[1+\cos 40^{\circ}-\left(\cos 120^{\circ}+\cos 40^{\circ}\right)\right] \\
& =4 \times \frac{3}{2}=6
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
f(x) & =9 \sin ^2 x-16 \cos ^2 x-10(3 \sin x-4 \cos x)-10(3 \sin x+4 \cos x)+100 \\
& =25 \sin ^2 x-60 \sin x+84 \\
& =(5 \sin x-6)^2+48
\end{aligned}
\)
The minimum value of \(f(x)\) occurs when \(\sin x=1\).
Therefore, the minimum value of \(\sqrt{f(x)}\) is 7.

Hint

(d)
\(
\begin{aligned}
& \text { In } \triangle A B C, \tan A+\tan B+\tan C=\tan A \tan B \tan C \\
& \Rightarrow x+x+1+1-x=x(1+x)(1-x) \\
& \Rightarrow 2+x=x-x^3 \\
& \Rightarrow x^3=-2 \Rightarrow x=-2^{1 / 3} \\
& \Rightarrow \tan A=x<0 \Rightarrow A \text { is obtuse } \\
& \Rightarrow \tan B=x+1=1-2^{1 / 3}<0
\end{aligned}
\)
Hence, \(A\) and \(B\) are obtuse, which is not possible in a triangle.
Hence, no such triangle can exist.

Hint

(a)
\(
\begin{aligned}
& \text { Given } \log _{10}\left(\frac{\sin 2 x}{2}\right)=-1 \\
& \Rightarrow \frac{\sin 2 x}{2}=\frac{1}{10} \\
& \Rightarrow \sin 2 x=\frac{1}{5} \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Also } \log _{10}(\sin x+\cos x)=\frac{\log _{10}\left(\frac{n}{10}\right)}{2} \\
& \Rightarrow \log _{10}(\sin x+\cos x)^2=\log _{10}\left(\frac{n}{10}\right) \\
& \Rightarrow 1+\sin 2 x=\frac{n}{10} \\
& \Rightarrow 1+\frac{1}{5}=\frac{n}{10} \\
& \Rightarrow \frac{6}{5}=\frac{n}{10} \\
& \Rightarrow \frac{n}{3}=4
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
\frac{2 \sin 4^{\circ} \cos 3^{\circ}+2 \sin 4^{\circ} \cos 1^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \sin 4^{\circ}} & =\frac{2 \sin 4^{\circ}\left[\cos 3^{\circ}+\cos 1^{\circ}\right]}{\cos 1^{\circ} \cos 2^{\circ} \sin 4^{\circ}} \\
& =\frac{4 \cos 2^{\circ} \cos 1^{\circ}}{\cos 1^{\circ} \cos 2^{\circ}}=4
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{1}{16} \\
& \Rightarrow\left(\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right) \sin \frac{C}{2}=\frac{1}{16} \\
& \Rightarrow \sin \frac{C}{2}\left(\frac{1}{2}-\sin \frac{C}{2}\right)=\frac{1}{16} \\
& \Rightarrow \sin ^2 \frac{C}{2}-\frac{1}{2} \sin \frac{C}{2}+\frac{1}{16}=0 \\
& \Rightarrow\left(\frac{1}{4}-\sin \frac{C}{2}\right)^2=0 \\
& \Rightarrow \sin \frac{C}{2}=\frac{1}{4} \\
& \Rightarrow \cos C=1-2 \sin ^2 \frac{C}{2}=1-\frac{1}{8}=\frac{7}{8}
\end{aligned}
\)
\(
8 \cos C=7
\)

Hint

(b)

\(
\begin{aligned}
& \text { In the triangle, } \\
& \tan A+\tan B+\tan C=\tan A \tan B \tan C \\
& \Rightarrow \frac{1}{2} \frac{(2 k+1)}{2} \frac{(4 k+1)}{2}=\frac{3}{2}+3 k \\
& \Rightarrow \frac{8 k^2+6 k+1}{8}=\frac{3+6 k}{2} \\
& \Rightarrow 8 k^2+6 k+1=12+24 k \\
& \Rightarrow 8 k^2-18 k-11=0 \\
& \Rightarrow 8 k^2-22 k+4 k-11=0 \\
& \Rightarrow(2 k+1)(4 k-11)=0 \\
& \Rightarrow k=-1 / 2 \text { or } 11 / 4 \\
& \text { For } k=-1 / 2, \tan B=0 \text { (not possible) } \\
& \therefore k=11 / 4
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& 4 \sin ^3 x \cos 3 x+4 \cos ^3 x \sin 3 x=\frac{3}{2} \\
& \Rightarrow(3 \sin x-\sin 3 x) \cos 3 x+(3 \cos x+\cos 3 x) \sin 3 x=\frac{3}{2} \\
& \Rightarrow 3[\sin x \cos 3 x+\cos x \sin 3 x]=\frac{3}{2} \\
& \Rightarrow \sin 4 x=\frac{1}{2}
\end{aligned}
\)
\(
8 \sin 4 x=4
\)

Hint

(a)
\(
\begin{aligned}
& \sin ^3 x+p^3+1=3 p \sin x \\
& \Rightarrow(\sin x+p+1)\left(\sin ^2 x+1+p^2-\sin x-p-p \sin x\right)=0
\end{aligned}
\)
Therefore, either \(\sin x+p+1=0 \Rightarrow p=-(1+\sin x)\), or \(\sin x=1=p\)
Hence, only one value of \(p(p>0)\) is possible which is given by \(p=1\).

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
& |\sin x \cos x|+|\tan x+\cot x|=\sqrt{3} \\
& \Rightarrow|\sin x \cos x|+\frac{1}{|\sin x \cos x|}=\sqrt{3} \\
& |\sin x \cos x|+\frac{1}{|\sin x \cos x|} \geq 2
\end{aligned}\\
&\text { Hence, there is no solution. }
\end{aligned}
\)

Hint

(c) The range of \(\cos x\) is \([-1,1]\).
The range of \(\sin x\) is \([-1,1]\).
We know that \(-1 \leq \cos x \leq 1\).
So, \(2 \leq 3+\cos x \leq 4\).
Therefore, \(4 \leq(3+\cos x)^2 \leq 16\).
We know that \(0 \leq \sin ^2 x \leq 1\).
So, \(0 \leq \sin ^8 x \leq 1\).
This implies \(0 \leq 2 \sin ^8 x \leq 2\).
Therefore, \(-2 \leq-2 \sin ^8 x \leq 0\).
Adding 4 to all parts, \(2 \leq 4-2 \sin ^8 x \leq 4\).
For the equation to hold, both sides must be equal to the common value in their ranges.
The only common value is 4.
So, \((3+\cos x)^2=4\) and \(4-2 \sin ^8 x=4\).
From \((3+\cos x)^2=4\), we get \(3+\cos x= \pm 2\).
Since \(3+\cos x \geq 2\), we must have \(3+\cos x=2\).
This gives \(\cos x=-1\).
From \(4-2 \sin ^8 x=4\), we get \(-2 \sin ^8 x=0\).
This implies \(\sin ^8 x=0[latex], so [latex]\sin x=0[latex].
We need [latex]\cos x=-1\) and \(\sin x=0\).
This occurs when \(x\) is an odd multiple of \(\pi\).
So, \(x=(2 n+1) \pi\) for integer \(n\).
The interval is \([0,5 \pi]\).
For \(n=0, x=\pi\).
For \(n=1, x=3 \pi\).
For \(n=2, x=5 \pi\).
The number of roots of the equation in the interval \([0,5 \pi]\) is 3.

Hint

(d)

\(
\begin{aligned}
&\begin{aligned}
& \frac{\sin x}{\cos 3 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\sin 9 x}{\cos 27 x}=0 \\
& \Rightarrow \frac{2 \sin x \cos x}{2 \cos 3 x \cos x}+\frac{2 \sin 3 x \cos 3 x}{2 \cos 9 x \cos 3 x}+\frac{2 \sin 9 x \cos 9 x}{2 \cos 27 x \cos 9 x}=0 \\
& \Rightarrow \frac{\sin (3 x-x)}{2 \cos 3 x \cos x}+\frac{\sin (9 x-3 x)}{2 \cos 9 x \cos 3 x}+\frac{\sin (27 x-9 x)}{2 \cos 27 x \cos 9 x}=0 \\
& \Rightarrow(\tan 3 x-\tan x)+(\tan 9 x-\tan 3 x)+(\tan 27 x-\tan 9 x)=0 \\
& \Rightarrow \tan 27 x-\tan x=0 \\
& \Rightarrow \tan x=\tan 27 x \\
& \Rightarrow 27 x=n \pi+x, n \in I \\
& \Rightarrow x=\frac{n \pi}{26}, n \in I \\
& \Rightarrow x=\frac{\pi}{26}, \frac{2 \pi}{26}, \frac{3 \pi}{26}, \frac{4 \pi}{26}, \frac{5 \pi}{26}, \frac{6 \pi}{26}
\end{aligned}\\
&\text { Hence, there are six solutions. }
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& (\sqrt{3}+1)^{2 x}+(\sqrt{3}-1)^{2 x}=2^{3 x}=(2 \sqrt{2})^{2 x} \\
& \Rightarrow\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^{2 x}+\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2 x}=1 \\
& \Rightarrow\left(\sin 75^{\circ}\right)^{2 x}+\left(\cos 75^{\circ}\right)^{2 x}=1 \\
& \Rightarrow x=1
\end{aligned}
\)

Hint

(d)
The equation is \(\sin x-\sqrt{3} \cos x=\frac{4 m-6}{4-m}\).
The solutions \(x\) are in the interval \([0,2 \pi]\).
The range of \(a \sin x+b \cos x\) is \(\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]\).
Step 1: Transform the left side of the equation.
Multiply and divide \(\sin x-\sqrt{3} \cos x\) by \(\sqrt{1^2+(-\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2\).
\(
2\left(\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x\right)
\)
Use the identity \(\sin (A-B)=\sin A \cos B-\cos A \sin B\).
\(
2\left(\sin x \cos \left(\frac{\pi}{3}\right)-\cos x \sin \left(\frac{\pi}{3}\right)\right)=2 \sin \left(x-\frac{\pi}{3}\right)
\)
Step 2: Determine the range of the left side.
The range of \(\sin \left(x-\frac{\pi}{3}\right)\) is \([-1,1]\).
The range of \(2 \sin \left(x-\frac{\pi}{3}\right)\) is \([-2,2]\).
Step 3: Set up the inequality for the right side.
Since the equation has solutions, the right side must be within the range of the left side.
\(
-2 \leq \frac{4 m-6}{4-m} \leq 2
\)
Step 4: Solve the inequality for \(m\).
Consider two separate inequalities:
\(
-2 \leq \frac{4 m-6}{4-m} \Longrightarrow-2(4-m) \leq 4 m-6 \Longrightarrow-8+2 m \leq 4 m-6 \Longrightarrow-2 \leq 2 m \Longrightarrow-1 \leq m
\)
\(
\frac{4 m-6}{4-m} \leq 2 \Longrightarrow 4 m-6 \leq 2(4-m) \Longrightarrow 4 m-6 \leq 8-2 m \Longrightarrow 6 m \leq 14 \Longrightarrow m \leq \frac{14}{6} \Longrightarrow m \leq \frac{7}{3}
\)
Combine the inequalities: \(-1 \leq m \leq \frac{7}{3}\).
Step 5: Find the integral values of \(m\).
The integral values of \(m\) satisfying \(-1 \leq m \leq \frac{7}{3}[latex] are [latex]-1,0,1,2\).
The number of integral values is 4.
The number of integral values of \(m\) is 4.

Hint

(a)

\(
\begin{aligned}
&\text { Adding given equations, we get }\\
&\begin{aligned}
& 2=\frac{3 a}{2}+\frac{a^2}{2} \\
& \Rightarrow a^2+3 a-4=0 \\
& \Rightarrow(a+4)(a-1)=0 \\
& \Rightarrow a=1(\text { as } a=-4 \text { is rejected })
\end{aligned}
\end{aligned}
\)

Explanation: The first equation is \(\sin ^2 x+\cos ^2 y=\frac{3 a}{2}\).
The second equation is \(\cos ^2 x+\sin ^2 y=\frac{a^2}{2}\).
The range of \(\sin ^2 \theta\) and \(\cos ^2 \theta\) is \([0,1]\).
The identity \(\sin ^2 \theta+\cos ^2 \theta=1\) is fundamental.
Step 1: Add the two equations.
Add the left-hand sides: \(\left(\sin ^2 x+\cos ^2 y\right)+\left(\cos ^2 x+\sin ^2 y\right)\).
Add the right-hand sides: \(\frac{3 a}{2}+\frac{a^2}{2}\).
The sum is \(\sin ^2 x+\cos ^2 y+\cos ^2 x+\sin ^2 y=\frac{3 a}{2}+\frac{a^2}{2}\).
Step 2: Simplify the sum using trigonometric identities.
Rearrange terms: \(\left(\sin ^2 x+\cos ^2 x\right)+\left(\sin ^2 y+\cos ^2 y\right)=\frac{a^2+3 a}{2}\).
Apply \(\sin ^2 \theta+\cos ^2 \theta=1: 1+1=\frac{a^2+3 a}{2}\).
This simplifies to \(2=\frac{a^2+3 a}{2}\).
Step 3: Solve the resulting quadratic equation for \(a\).
Multiply by 2: \(4=a^2+3 a\).
Rearrange into standard quadratic form: \(a^2+3 a-4=0\).
Factor the quadratic equation: \((a+4)(a-1)=0\).
The possible values for \(a\) are \(a=-4\) or \(a=1\).
Step 4: Check the validity of \(a\) values based on the range of \(\sin ^2 \theta\) and \(\cos ^2 \theta\).
For the first equation, \(\sin ^2 x+\cos ^2 y=\frac{3 a}{2}\).
Since \(0 \leq \sin ^2 x \leq 1\) and \(0 \leq \cos ^2 y \leq 1\), then \(0 \leq \sin ^2 x+\cos ^2 y \leq 2\).
So, \(0 \leq \frac{3 a}{2} \leq 2\), which implies \(0 \leq 3 a \leq 4\), or \(0 \leq a \leq \frac{4}{3}\).
For the second equation, \(\cos ^2 x+\sin ^2 y=\frac{a^2}{2}\).
Similarly, \(0 \leq \frac{a^2}{2} \leq 2\), which implies \(0 \leq a^2 \leq 4\), or \(-2 \leq a \leq 2\).
Combining all conditions, a must satisfy \(0 \leq a \leq \frac{4}{3}\) and \(-2 \leq a \leq 2\).
The intersection of these ranges is \(0 \leq a \leq \frac{4}{3}\).
From the possible values \(a=-4\) and \(a=1\), only \(a=1\) falls within this valid range.
The value of \(a\) for which the system of equations has a solution is 1.

Hint

(c)

\(
\begin{aligned}
\cos 4 x & =2 \cos ^2 2 x-1 \\
& =2\left(2 \cos ^2 x-1\right)^2-1 \\
& =2\left(4 \cos ^4 x+1-4 \cos ^2 x\right)-1 \\
& =8 \cos ^4 x-8 \cos ^2 x+1 \\
\therefore a_0 & =1 ; a_1=-8, a_2=8 \\
\therefore 5 a_0 & +a_1+a_2=5
\end{aligned}
\)

Hint

(b)
\(
\begin{aligned}
& 1-\sin ^2 x-\sin x+a=0 \\
& \Rightarrow \sin ^2 x+\sin x-(a+1)=0
\end{aligned}
\)
From Eq. (i), we get
\(
\sin ^2 x+\sin x=(a+1) \dots(i)
\)
For \(x \in(0, \pi / 2)\), the range of \(\sin ^2 x+\sin x\) is \((0,2)\).
\(
\Rightarrow 0<(a+1)<2 \Rightarrow a \in(-1,1)
\)

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& a \sin x+1-2 \sin ^2 x=2 a-7 \\
& \Rightarrow 2 \sin ^2 x-a \sin x+(2 a-8)=0 \\
& \Rightarrow \sin x=\frac{a \pm \sqrt{a^2-8(2 a-8)}}{4}=\frac{a \pm(a-8)}{4}=2 \text { or } \frac{a-4}{2}
\end{aligned}\\
&\text { For a solution }-1 \leq \frac{a-4}{2} \leq 1 \text {, we have } 2 \leq a \leq 6 \text {. }
\end{aligned}
\)

Hint

(a)

\(
\frac{\sin ^2\left(x-\frac{\pi}{4}\right)}{\cos 2 x}=\frac{\frac{1}{2}(\sin x-\cos x)^2}{\cos ^2 x-\sin ^2 x}=\frac{-\frac{1}{2}(\sin x-\cos x)}{\cos x+\sin x}=-\frac{1}{2} \tan \left(x-\frac{\pi}{4}\right)
\)
\(
\text { Given equation reduces to } 2^{\tan \left(x-\frac{\pi}{4}\right)}-2(0.25)^{\frac{1}{2} \tan \left(x-\frac{\pi}{4}\right)}+1=0
\)
Let \(y=2^{\tan \left(x-\frac{\pi}{4}\right)}\).
The equation is \(y-2 y+1=0\).
This simplifies to \(-y+1=0\), so \(y=1\).
\(
\Rightarrow 2^{\tan \left(x-\frac{\pi}{4}\right)}=1
\)
\(\Rightarrow x=\pi / 4\) which is not possible as \(\cos 2 x=0\) for this value of \(x\), which is not defining the original equation.

Hint

(d)

\(
\begin{aligned}
& \sin ^4 x-\cos ^2 x \sin x+2 \sin ^2 x+\sin x=0 \\
& \Rightarrow \sin x\left[\sin ^3 x-\cos ^2 x+2 \sin x+1\right]=0
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& \Rightarrow \sin x\left[\sin ^3 x+\sin ^2 x+2 \sin x\right]=0 \\
& \Rightarrow \sin ^2 x\left[\sin ^2 x+\sin x+2\right]=0 \\
& \Rightarrow \sin x=0, \text { where } x=0, \pi, 2 \pi, 3 \pi
\end{aligned}\\
&\text { Hence, there are four solutions. }
\end{aligned}
\)

Explanation: The number of solutions for the given trigonometric equation in the interval \(0 \leq x \leq 3 \pi\).
The equation is \(\sin ^4 x-\cos ^2 x \sin x+2 \sin ^2 x+\sin x=0\).
The interval for \(x\) is \(0 \leq x \leq 3 \pi\).
How to solve?
Factor out \(\sin x\) and use trigonometric identities to simplify the equation, then solve for \(x\) in the given interval.
Step 1: Factor out \(\sin x\) from the equation.
The equation becomes \(\sin x\left(\sin ^3 x-\cos ^2 x+2 \sin x+1\right)=0\).
Step 2: Use the identity \(\cos ^2 x=1-\sin ^2 x\).
Substitute \(\cos ^2 x\) in the factored equation: \(\sin x\left(\sin ^3 x-\left(1-\sin ^2 x\right)+2 \sin x+1\right)=0\).
Simplify the expression inside the parenthesis: \(\sin x\left(\sin ^3 x+\sin ^2 x+2 \sin x\right)=0\).
Step 3: Factor out \(\sin x\) again from the expression inside the parenthesis.
The equation becomes \(\sin x \cdot \sin x\left(\sin ^2 x+\sin x+2\right)=0\).
This simplifies to \(\sin ^2 x\left(\sin ^2 x+\sin x+2\right)=0\).
Step 4: Solve for \(\sin x\).
This implies either \(\sin ^2 x=0\) or \(\sin ^2 x+\sin x+2=0\).
For \(\sin ^2 x=0\), we have \(\sin x=0\).
For \(\sin ^2 x+\sin x+2=0\), let \(y=\sin x\). The quadratic equation is \(y^2+y+2=0\).
Calculate the discriminant \(\Delta=b^2-4 a c=1^2-4(1)(2)=1-8=-7\).
Since the discriminant is negative, there are no real solutions for \(y\) in \(y^2+y+2=0\).
Step 5: Find the values of \(x\) for \(\sin x=0\) in the interval \(0 \leq x \leq 3 \pi\).
The values are \(x=0, x=\pi, x=2 \pi\), and \(x=3 \pi\).
There are 4 solutions to the equation in the given interval.