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
$$ \begin{aligned} & 5 \frac{1-y}{1+y}-\frac{1+y}{2}=2 y+9 \\ & \Rightarrow \quad 5\left(2-2 y-1-y^{2}-2 y\right)=2(1+y)(2 y+9) \\ & \Rightarrow \quad 5\left(1-4 y-y^{2}\right)=2\left(2 y+9+2 y^{2}+9 y\right) \\ & \Rightarrow \quad 5-20 y-5 y^{2}=22 y+18+4 y^{2} \\ & \Rightarrow \quad 9 y^{2}+42 y+13=0 \\ & \Rightarrow \quad 9 y^{2}+3 y+39 y+13=0 \\ & \Rightarrow 3 y(3 y+1)+13(3 y+1)=0 \\ & \Rightarrow \quad(3 y+1)(3 y+13)=0 \\ & \Rightarrow \quad y=-\frac{1}{3},-\frac{13}{3} \\ & \therefore \quad \cos 2 x=-\frac{1}{3},-\frac{13}{3} \\ & \therefore \quad \cos 2 x=-\frac{1}{3} \quad \because \cos 2 x \neq-\frac{13}{3} \end{aligned} $$
| $\Rightarrow$ | $2 \cos 2 x \cos x+2 \cos 3 x \cos x=0$ |
|---|---|
| $\Rightarrow$ | $2 \cos x(\cos 2 x+\cos 3 x)=0$ |
| $\Rightarrow$ | $2 \cos x 2 \cos \frac{5 x}{2} \cos \frac{x}{2}=0$ |
| $\cos x \cdot \cos \frac{5 x}{2} \cdot \cos \frac{x}{2}=0$ | |
| c | $\begin{aligned} (\cos x & =0 \ \cos \frac{5 x}{2} & =0\end{aligned}$) |
| $\cos \frac{x}{2}=0$ | |
| ow, | $\cos x=0$ |
| $\begin{aligned}(x & =\frac{\pi}{2}, \frac{3 \pi}{2} \quad[\because 0 \leq x<2 \pi] \ \cos \frac{5 x}{2} & =0\end{aligned}$) | |
| $\rightarrow$ | $\frac{5 x}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \frac{9 \pi}{2}, \frac{11 \pi}{2} \ldots$ |
| $x=\frac{\pi}{5}, \frac{3 \pi}{5}, \pi, \frac{7 \pi}{5}, \frac{9 \pi}{5} \quad[\because 0 \leq x<2 \pi]$ | |
| and | $\cos \frac{x}{2}=0$ |
| $\Rightarrow$ | $\frac{x}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \ldots$ |
| $\Rightarrow$ | $[\because 0 \leq x<2 \pi]$ |
| ence, | $x=\frac{\pi}{2}, \frac{3 \pi}{2}, \pi, \frac{\pi}{5}, \frac{3 \pi}{5}, \frac{7 \pi}{5}, \frac{9 \pi}{5}$ |
Now, $\quad \cos 4 x=2 \cos ^{2} 2 x-1$
$$ =2-\frac{1}{3}^{2}-1=\frac{2}{9}-1=-\frac{7}{9} $$
