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

Question

If e^( cos^2 ⁡x+ cos^4 ⁡x+ cos^6 ⁡x+ … ) log_e ⁡2 satisfies the equation t^2-9t+8=0, then the value of 2 sin ⁡x sin ⁡x+ 3 cos ⁡x(0<x< 2) is:

Accepted Answer

Correct answer: C. c JEE Main 2021 e^( cos^2 ⁡x+ cos^4 ⁡x+ … … ) ln ⁡2=2^ cos^2 ⁡x+ cos^4 ⁡x+ … … =2^ cot^2x t^2-9t+8=0 ⇒t=1 ,8 ⇒2^ cot^2x=1 ,8 ⇒ cot^2 ⁡x=0 ,3 0 ⇒ 2 sin ⁡x sin ⁡x+ 3 cos ⁡x= 21+ 3 cot ⁡x= 24= 12

$If e^{(cos^{2} x + cos^{4} x + cos^{6} x + \dots \infty) log_{e} 2} satisfies the equation t^{2} - 9 t + 8 = 0, then the value of \frac{2 sin x}{sin x + 3 cos x} (0 < x < \frac{π}{2}) is:$

Solution

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