If e^(^2x+^4x+^6x+)2 satisfies the equation t^2-9t+8=0, then the value of 2 x x+√3 x, 0

Question

If $e^{(\cos^2x+\cos^4x+\cos^6x+\cdots\infty)\ln2}$ satisfies the equation $t^2-9t+8=0$, then the value of $\dfrac{2\sin x}{\sin x+\sqrt3\cos x}$, $0<x<\dfrac{\pi}{2}$, is:

Accepted Answer

$\cos^2x+\cos^4x+\cdots = \dfrac{\cos^2x}{1-\cos^2x} = \cot^2x$, so the expression is $2^{\cot^2x}$.

$t^2-9t+8=0 \Rightarrow t=1,8$. Since $2^{\cot^2x}\ge1$ automatically, both are possible, but $t=1$ gives $\cot^2x=0$ and $t=8$ gives $\cot^2x=3$.

Taking $\cot^2x=3$ (i.e. $\cot x=\sqrt3$):
$$\frac{2\sin x}{\sin x+\sqrt3\cos x} = \frac{2}{1+\sqrt3\cot x} = \frac{2}{4} = \frac12.$$

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