Trigonometric equations

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The number of solutions of the equation $\tan x + \sec x = 2 \cos x, x \in [0, 2 π]$ is

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Solution

$\tan x + \sec x = 2 \cos x$
$\begin{array}{l} \Rightarrow 1 + \sin x = 2 \cos^{2} x = 2 (1 - \sin^{2} x) \\ \Rightarrow (1 + \sin x) (1 - 2 + 2 \sin x) = 0 \\ \Rightarrow \sin x = - 1 or \sin x = 1 / 2 \end{array}$
But $\sin x = - 1 \Rightarrow \cos x = 0$ This is not possible.
Thus $\sin x = 1 / 2 \Rightarrow x = π / 6, 5 π / 6, \in [0, 2 π]$

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Trigonometric equations

Remember concepts with our Masterclasses.

The number of solutions of the equation tan⁡x+sec⁡x=2cos⁡x,x∈[0,2π] is

tan⁡x+sec⁡x=2cos⁡x⇒ 1+sin⁡x=2cos2⁡x=21−sin2⁡x⇒ (1+sin⁡x)(1−2+2sin⁡x)=0⇒ sin⁡x=−1 or sin⁡x=1/2But sin⁡x=−1⇒cos⁡x=0 This is not possible.Thus sin⁡x=1/2⇒x=π/6,5π/6,∈[0,2π]

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The number of solutions of the equation $\tan x + \sec x = 2 \cos x, x \in [0, 2 π]$ is