Trigonometrical Equations 1 Question 26
26. If $e^{\left[\left(\sin ^2 x+\sin ^4 x+\sin ^6 x+\ldots+\infty\right) \log _e 2\right]}$ satisfies the equation $x^2-9 x+8=0$ ,then the value of $g(x)=\frac{\cos x}{\cos x+\sin x}$ is.
(1991, 4M)
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Answer:
Correct Answer: 26. $\theta=n \pi$ or $n \pi+(-1)^{n} \frac{\pi}{6}$
Solution:
- $e^{\left[\left(\sin ^2 x+\sin ^4 x+\sin ^6 x+\ldots+\infty\right) \log _e 2\right]}$
$$ \begin{aligned} & =e^{\frac{\sin ^{2} x}{1-\sin ^{2} x} \cdot \log _e 2}=e^{\log _e 2 \frac{\sin ^{2} x}{\cos ^{2} x}} \\ & \Rightarrow \quad 2^{\tan ^{2} x} \text { satisfies } x^{2}-9 x+8=0 \\ & \Rightarrow \quad x=1,8 \\ & \therefore \quad 2^{\tan ^{2} x}=1 \quad \text { and } \quad 2^{\tan ^{2} x}=8 \\ & \Rightarrow \quad \tan ^{2} x=0 \quad \text { and } \quad \tan ^{2} x=3 \\ & \Rightarrow \quad x=n \pi \quad \text { and } \quad \tan ^{2} x=\tan \frac{\pi^{2}}{3} \\ & \Rightarrow \quad x=n \pi \text { and } \quad x=n \pi \pm \frac{\pi}{3} \end{aligned} $$
Neglecting $x=n \pi$ as $0<x<\frac{\pi}{2}$
$$ \begin{aligned} & \Rightarrow \quad x=\frac{\pi}{3} \in 0, \frac{\pi}{2} \\ & \therefore \quad \frac{\cos x}{\cos x+\sin x}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\sqrt{3}}{2}}=\frac{1}{1+\sqrt{3}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\ & \Rightarrow \quad \frac{\cos x}{\cos x+\sin x}=\frac{\sqrt{3}-1}{2} \end{aligned} $$
