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Trigonometric Equations

Section: Mathematics

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Question : 9 of 24

Marks: +1, -0

Number of solutions of the equation

\cos\theta + \cos 2\theta - \sqrt{3}(\sin\theta + \sin 2\theta) + 1 = 0

lying in the interval

(0,2\pi)

[AP EAPCET 21 May 2025 S2]

3
6
5
4

Solution:

We have,

\cos\theta + \cos 2\theta - \sqrt{3}(\sin\theta + \sin 2\theta) + 1 = 0

\Rightarrow \cos\theta + 2\cos^2\theta - \sqrt{3}\sin\theta - 2\sqrt{3}\sin\theta\cos\theta = 0

\Rightarrow (\cos\theta - \sqrt{3}\sin\theta) + (2\cos^2\theta - 2\sqrt{3}\sin\theta\cos\theta) = 0

\Rightarrow 1(\cos\theta - \sqrt{3}\sin\theta) + 2\cos\theta(\cos\theta - \sqrt{3}\sin\theta) = 0

\Rightarrow (\cos\theta - \sqrt{3}\sin\theta)(2\cos\theta + 1) = 0

\Rightarrow \cos\theta - \sqrt{3}\sin\theta = 0 \text{ or } 2\cos\theta = -1

\tan\theta = \frac{1}{\sqrt{3}} \text{ or } \cos\theta = -\frac{1}{2} = \cos\frac{2\pi}{3}

\theta = n\pi + \frac{\pi}{6} \text{ or } \theta = 2n\pi + \frac{2\pi}{3}

\theta = \frac{\pi}{6}, \frac{7\pi}{6} \text{ or } \theta = \frac{2\pi}{3}, \frac{4\pi}{3}

\because

Total number of solution in the interval

(0,2\pi)

is 4 .

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