Trigonometrical Equations 1 Question 9
9. If $5\left(\tan ^{2} x-\cos ^{2} x\right)=2 \cos 2 x+9$, then the value of $\cos 4 x$ is
(2017 Main)
(a) $-\frac{3}{5}$
(b) $\frac{1}{3}$
(c) $\frac{2}{9}$
(d) $-\frac{7}{9}$
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Answer:
Correct Answer: 9. (d)
Solution:
- Given, $5\left(\tan ^{2} x-\cos ^{2} x\right)=2 \cos 2 x+9$
$\Rightarrow \quad 5 \frac{2 \sin ^{2} x}{2 \cos ^{2} x}-\cos ^{2} x=2 \cos 2 x+9$
$\Rightarrow \quad 5 \frac{1-\cos 2 x}{1+\cos 2 x}-\frac{1+\cos 2 x}{2}=2 \cos 2 x+9$
Put $\cos 2 x=y$, we have
$ \begin{aligned} & 5 \frac{1-y}{1+y}-\frac{1+y}{2}=2 y+9 \\ & \Rightarrow \quad 5\left(2-2 y-1-y^{2}-2 y\right)=2(1+y)(2 y+9) \\ & \Rightarrow \quad 5\left(1-4 y-y^{2}\right)=2\left(2 y+9+2 y^{2}+9 y\right) \\ & \Rightarrow \quad 5-20 y-5 y^{2}=22 y+18+4 y^{2} \\ & \Rightarrow \quad 9 y^{2}+42 y+13=0 \\ & \Rightarrow \quad 9 y^{2}+3 y+39 y+13=0 \\ & \Rightarrow 3 y(3 y+1)+13(3 y+1)=0 \\ & \Rightarrow \quad(3 y+1)(3 y+13)=0 \\ & \Rightarrow \quad y=-\frac{1}{3},-\frac{13}{3} \\ & \therefore \quad \cos 2 x=-\frac{1}{3},-\frac{13}{3} \\ & \therefore \quad \cos 2 x=-\frac{1}{3} \quad \because \cos 2 x \neq-\frac{13}{3} \end{aligned} $
Now, $\quad \cos 4 x=2 \cos ^{2} 2 x-1$
$ =2-\frac{1}{3}^{2}-1=\frac{2}{9}-1=-\frac{7}{9} $
