⇒Trigonometric Functions

Correct answer option is (C)

$\begin{array}{ll}& 8{\cos }^2\theta +14\cos \theta +5=0\\ \therefore & 8{\cos }^2\theta +10\cos \theta +4\cos \theta +5=0\\ \therefore & 2\cos \theta (4\cos \theta +5)+1(4\cos \theta +5)=0\\ \therefore & (2\cos \theta +1)(4\cos \theta +5)=0\\ \therefore & \cos \theta =\frac{-1}{2}\text{ or }\cos \theta =\frac{-5}{4}\end{array}$

But $\cos \theta =\frac{-5}{4}$ is not possible as $\cos \theta \in [-1,1]$ for all values of $\theta$.

$\begin{array}{rl}& \therefore \cos \theta =\frac{-1}{2}\\ & \therefore \theta \in \{\frac{2\pi }{3},\frac{4\pi }{3}\}\end{array}$

Correct answer option is (A)

$\begin{array}{ll}& {\cos }^2\theta -2\sin \theta +\frac{1}{4}=0\\ \therefore & (1-{\sin }^2\theta )-2\sin \theta +\frac{1}{4}=0\\ \therefore & {\sin }^2\theta +2\sin \theta -\frac{5}{4}=0\\ \therefore & 4{\sin }^2\theta +8\sin \theta -5=0\\ \therefore & 4{\sin }^2\theta +10\sin \theta -2\sin \theta -5=0\\ \therefore & 2\sin \theta (2\sin \theta +5)-1(2\sin \theta +5)=0\\ \therefore & (2\sin \theta -1)(2\sin \theta +5)=0\\ \therefore & \sin \theta =\frac{1}{2}\text{ or }\sin \theta =\frac{-5}{2}\end{array}$

But $\sin \theta =\frac{-5}{2}$ is not possible as $\sin \theta \in [-1,1]$ for all values of $\theta$.

$\begin{array}{ll}\therefore & \sin \theta =\frac{1}{2}\\ \therefore & \sin \theta =\sin \frac{\pi }{6}\\ \therefore & \theta =\frac{n\pi }{1}+(-1{)}^n\frac{\pi }{6}\\ \therefore & A=1\text{ and }B=6\\ & \Rightarrow A+B=7\end{array}$

Correct answer option is (C)

$\begin{array}{rl}& \cos x+\cos y-\cos (x+y)=\frac{3}{2}\\ & \therefore 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-(2{\cos }^2\left(\frac{x+y}{2}\right)-1)=\frac{3}{2}\\ & \dots \left[\begin{array}{c}\because \cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)\text{ and }\\ \cos \theta =2{\cos }^2\left(\frac{\theta }{2}\right)-1\end{array}\right]\\ & \therefore 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-2{\cos }^2\left(\frac{x+y}{2}\right)=\frac{3}{2}-1\\ & \therefore 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-2{\cos }^2\left(\frac{x+y}{2}\right)=\frac{1}{2}\\ & \therefore 4{\cos }^2\left(\frac{x+y}{2}\right)-4\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)+1=0\end{array}$

Substituting $\cos \left(\frac{x+y}{2}\right)=\mathrm{t}$, we get

$4t^2-4t\cos \left(\frac{x-y}{2}\right)+1=0$

As $t$ is real, we get $b^2-4ac\ge 0$

$\begin{array}{rl}& \Rightarrow {[-4\cos \left(\frac{x-y}{2}\right)]}^2-4\times 4\times 1\ge 0\\ & \Rightarrow 16{\cos }^2\left(\frac{x-y}{2}\right)-16\ge 0\\ & \Rightarrow {\cos }^2\left(\frac{x-y}{2}\right)\ge 1\\ & \Rightarrow {\cos }^2\left(\frac{x-y}{2}\right)=1...[\because -1\le \cos \theta \le 1,\text{for all values of}\theta ]\\ & \Rightarrow \frac{x-y}{2}=0\\ & \Rightarrow x=y\end{array}$

Correct answer option is (B)

$\begin{array}{rl}& \frac{\tan 3x-1}{\tan 3x+1}=\sqrt{3}\\ \therefore & \frac{\tan 3x-\tan \frac{\pi }{4}}{1+(\tan 3x)(\tan \frac{\pi }{4})}=\sqrt{3}\\ \therefore & \tan (3x-\frac{\pi }{4})=\tan \left(\frac{\pi }{3}\right)\\ \therefore & 3x-\frac{\pi }{4}=n\pi +\frac{\pi }{3}\end{array}$

$\begin{array}{ll}\therefore & 3x=\mathrm{n}\pi +\frac{\pi }{3}+\frac{\pi }{4}\\ \therefore & 3x=\mathrm{n}\pi +\frac{7\pi }{12}\\ \therefore & x=\frac{\mathrm{n}\pi }{3}+\frac{7\pi }{36}\end{array}$

Comparing with $\frac{\mathrm{n}\pi }{\mathrm{p}}+\frac{7\pi }{\mathrm{q}}$, we get

$\begin{array}{rl}\mathrm{p}=3,\mathrm{q}& =36\\ \therefore \frac{\mathrm{p}}{\mathrm{q}}=\frac{3}{36}& =\frac{1}{12}\end{array}$

Correct answer option is (C)

$\begin{array}{rl}& \sin 7\theta +\sin \theta +\sin 4\theta =0\\ & \Rightarrow 2\sin 4\theta \cos 3\theta +\sin 4\theta =0\\ & \Rightarrow \sin 4\theta (2\cos 3\theta +1)=0\\ & \Rightarrow \sin 4\theta =0\text{ or }\cos 3\theta =\frac{-1}{2}\\ & \Rightarrow \sin 4\theta =0\text{ or }\cos 3\theta =\cos \left(\frac{2\pi }{3}\right)\\ & \Rightarrow 4\theta =n\pi \text{ or }3\theta =2n\pi \pm \frac{2\pi }{3}\\ & \Rightarrow \theta =\frac{n\pi }{4}\text{ or }\theta =\frac{2n\pi }{3}\pm \frac{2\pi }{9}\\ \therefore & \theta =\frac{2\pi }{9},\frac{\pi }{4},\frac{4\pi }{9},\frac{\pi }{2},\frac{3\pi }{4},\frac{8\pi }{9}\cdots [\because \theta \in (0,\pi )]\\ & \therefore \text{ Number of solutions }=6\end{array}$