The number of solutions of the equation log_(2sin⁡x)(1+cos⁡x)=2 in the interval 0,5π is

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MARKS Content Team · Accepted Answer

Correct answer: 3. log_(2sin⁡x)1+cos⁡x=2 1+cos⁡x=2sin⁡x^2=2sin^2⁡x 1+cos⁡x=2-2cos^2⁡x 2cos^2⁡x+cos⁡x-1=0 ⇒cos⁡x+12cos⁡x-1=0 ⇒cos⁡x=1/2 or cosx=-1 (rejected) ⇒x=(π)/3,(5π)/3 in 0,2π But at x=(5π)/3, √2sin⁡x is negative ⇒x=(π)/3 in 0,2π ⇒x=(π)/3, (π)/3+2π, (π)/3+4π ⇒3 solutions in 0,5π

The number of solutions of the equation $l o g_{\sqrt{2} \sin x} (1 + \cos x) = 2$ in the interval $[0, 5 π]$ is

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More Trigonometric Equations Questions

The number of solutions of the equation log2sin⁡x(1+cos⁡x)=2 in the interval 0,5π is

Solution

More Trigonometric Equations Questions

The number of solutions of the equation $l o g_{\sqrt{2} \sin x} (1 + \cos x) = 2$ in the interval $[0, 5 π]$ is