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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The number of solutions of the equation lo g _ 2 sin x ( 1 + cos x...

The number of solutions of the equation log2sinx(1+cosx)=2lo{g}_{\sqrt{2}sin x} ( 1 + cos x ) = 2 in the interval [0,5π]\left[0, 5\pi\right] is

Correct answer

3

Step-by-step solution

log⁡2sinx(1+cosx)=2\log_{\sqrt{2}sin x}\left(1 + cos x\right) = 2
1+cosx=(2sinx)2=2sin⁡2x1 + cos x = {\left(\sqrt{2}sin x\right)}^{2} = 2\sin^{2} x
1+cosx=2−2cos⁡2x1 + cos x = 2 - 2\cos^{2} x
2cos⁡2x+cosx−1=02\cos^{2} x + cos x - 1 = 0
⇒(cosx+1)(2cosx−1)=0⇒ \left(cos x + 1\right)\left(2cos x - 1\right) = 0
⇒cosx=12⇒ cos x = \dfrac{1}{2} or \cos x = - 1 (rejected)
⇒x=π3,5π3in[0,2π]⇒ x = \dfrac{\pi}{3}, \dfrac{5\pi}{3} in \left[0,2\pi\right]
But at x=5π3,2sinxx = \dfrac{5\pi}{3}, \sqrt{2}sin x is negative
⇒x=π3⇒ x = \dfrac{\pi}{3} in \left[0,2\pi\right]<br/><br />⇒ x = \dfrac{\pi}{3}, \dfrac{\pi}{3} + 2\pi, \dfrac{\pi}{3} + 4\pi<br/><br />⇒ 3solutionsin[0,5π]solutions in \left[0, 5\pi\right]
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