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.