Question

if sin^2 theta - 2 cos theta + 1/4 = 0 then the general value of theta is​

Answer 1

Answer

The general solution of the given equation is :

2nπ ± π/3 , n € Z

Given

The trigonometric equation is

$\bullet \: \: \sf \sin^{2}\theta - 2\cos\theta + \dfrac{1}{4} = 0$

To Find

• The general value of $\sf \theta$

Solution

Given ,

$\sf \sin {}^{2} \theta - 2 \cos \theta + \dfrac{1}{4} = 0 \\ \\ \sf\implies1 - \cos {}^{2} \theta - 2 \cos \theta + \dfrac{1}{4} = 0 \\ \\ \sf \implies4 - 4 \cos {}^{2} \theta - 8 \cos \theta + 1 = 0 \\ \\ \sf \implies - 4 \cos {}^{2} \theta - 8 \cos \theta + 5 = 0 \\ \\ \sf \implies4 \cos {}^{2} \theta + 8\cos \theta - 5 = 0 \\ \\ \sf \sf\implies4 { \cos}^{2} \theta - 2 \cos \theta + 10 \cos \theta - 5 = 0 \\ \\ \sf \implies2 \cos \theta( \cos \theta - 1) + 5(2 \cos \theta - 1) = 0 \\ \\ \sf \implies( 2\cos \theta - 1)(2 \cos \theta + 5) = 0$

Now we have ,

$\sf\implies 2\cos\theta - 1 = 0 \: \: and \: \: implies 2\cos\theta + 5 = 0 \\\\ \sf\implies \cos\theta = \dfrac{1}{2} \: \: and \: \implies \cos\theta = \dfrac{-5}{2} (not \: possible )$

So we are taking ,

$\sf\implies \cos\theta = \dfrac{1}{2} \\\\ \sf\implies \cos\theta = \cos\dfrac{\pi}{3} \\\\ \sf\implies x = 2n\pi \pm \dfrac{\pi}{3} , n \in Z$