Question

How do you solve $ 4{\sin ^2}x - 1 = 0 $ in the interval $ 0 \leqslant x \leqslant 2\pi $ ?

Answer 1

Hint: In the given question, we are required to find all the possible values of x that satisfy the given trigonometric equation $ 4{\sin ^2}x - 1 = 0 $ in the interval $ 0 \leqslant x \leqslant 2\pi $ . For solving such types of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.

Complete step-by-step answer:
We have to solve the given trigonometric equation $ 4{\sin ^2}x - 1 = 0 $ . We know that $ \cos \left( {2\theta } \right) = 1 - 2{\sin ^2}\theta $ . So, we can convert the $ {\sin ^2}\theta $ term into $ \cos 2\theta $ using the double angle formula of cosine.
So, we can solve for the value of $ {\sin ^2}\theta $ in $ \cos \left( {2\theta } \right) = 1 - 2{\sin ^2}\theta $ and then substitute the value in $ 4{\sin ^2}x - 1 = 0 $ so as to obtain a trigonometric equation in $ \cos \left( {2x} \right) $ that can be easily solved.
So, we get $ {\sin ^2}x = \left( {\dfrac{{1 - \cos 2x}}{2}} \right) $ using the double angle formula of cosine $ \cos \left( {2\theta } \right) = 1 - 2{\sin ^2}\theta $ . Now, we have,
 $ \Rightarrow 4{\sin ^2}x - 1 = 0 $
 $ \Rightarrow 4\left( {\dfrac{{1 - \cos 2x}}{2}} \right) - 1 = 0 $
Simplifying further,
 $ \Rightarrow 2 - 2\cos 2x - 1 = 0 $
\[ \Rightarrow 2\cos 2x = 1\]
\[ \Rightarrow \cos 2x = \left( {\dfrac{1}{2}} \right)\]
We know that the general solution for the equation $ \cos \left( \theta \right) = \cos \left( \phi \right) $ is $ \theta = 2n\pi \pm \phi $ . So, first we have to convert $ \cos 2x = \dfrac{1}{2} $ into $ \cos \left( \theta \right) = \cos \left( \phi \right) $ form.
Now, we know that the value of \[\cos \left( {\dfrac{\pi }{3}} \right)\] is $ \dfrac{1}{2} $ .
So, we get the trigonometric equation as \[\cos 2x = \cos \left( {\dfrac{\pi }{3}} \right)\].
Hence, for \[\cos 2x = \cos \left( {\dfrac{\pi }{3}} \right)\] , we have $ 2x = 2n\pi \pm \dfrac{\pi }{3} $ .
Dividing both sides of the equation $ 2x = 2n\pi \pm \dfrac{\pi }{3} $ by $ 2 $ , we get,
 $ \Rightarrow x = n\pi \pm \dfrac{\pi }{6} $
So the possible values of x for $ 4{\sin ^2}x - 1 = 0 $ are $ x = n\pi \pm \dfrac{\pi }{6} $ where n is any integer.
Now, we have to find the solution for the trigonometric equation in the interval $ 0 \leqslant x \leqslant 2\pi $ .
So, we substitute the values of n as $ 0 $ , $ 1 $ , $ 2 $ in $ x = n\pi \pm \dfrac{\pi }{6} $ .
So, putting $ n = 0 $ , we get the value of x as $ x = \left( 0 \right)\pi \pm \dfrac{\pi }{6} = \pm \dfrac{\pi }{6} $
So, putting $ n = 1 $ , we get the value of x as $ x = \left( 1 \right)\pi \pm \dfrac{\pi }{6} = \pi \pm \dfrac{\pi }{6} = \dfrac{{7\pi }}{6},\dfrac{{5\pi }}{6} $
So, putting $ n = 2 $ , we get the value of x as $ x = \left( 2 \right)\pi \pm \dfrac{\pi }{6} = 2\pi \pm \dfrac{\pi }{6} = \dfrac{{11\pi }}{6},\dfrac{{13\pi }}{6} $
Therefore, the solutions for value of x satisfying the trigonometric equation $ 4{\sin ^2}x - 1 = 0 $ and lying in the interval $ 0 \leqslant x \leqslant 2\pi $ are $ \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6} $ .
So, the correct answer is “ $ \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6} $ . ”.

Note: Such trigonometric equations can be solved by various methods by applying suitable trigonometric identities and formulae. The general solution of a given trigonometric solution may differ in form, but actually represents the correct solutions. The different forms of general equations are interconvertible into each other.