Question

How do you find all the solutions of $2{\cos ^2}x - \sin x - 1 = 0$

Answer 1

Hint:In the question, we are actually given a polynomial equation containing two one variable quantity that is “x” but it is written as two trigonometric functions so we first convert the equation in terms of one trigonometric function, now it is a function of either $\cos x$ or $\sin x$ . It becomes easier to solve the given polynomial equation if we replace the trigonometric function in the given equation with any other variable. We know that the number of roots of a polynomial equation is equal to its degree, so the given equation will have two roots that can be found by an appropriate method. So, we can find the values of x after finding the possible values of the trigonometric function.

Complete step by step answer:
We have to solve the equation $2{\cos ^2}x - \sin x - 1 = 0$

We know that –
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x \\
$

Using this value in the given equation, we get –
$
2(1 - {\sin ^2}x) - \sin x - 1 = 0 \\

\Rightarrow 2 - 2{\sin ^2}x - \sin x - 1 = 0 \\

\Rightarrow - 2{\sin ^2}x - \sin x + 1 = 0 \\

\Rightarrow 2{\sin ^2}x + \sin x - 1 = 0 \\
$

Let $\sin x = t$
$ \Rightarrow 2{t^2} + t - 1 = 0$

We can solve the above equation by factorization to find the value of t, as follows –
$

2{t^2} + 2t - t - 1 = 0 \\

2t(t + 1) - 1(t + 1) = 0 \\

\Rightarrow (2t - 1)(t + 1) = 0 \\

\Rightarrow 2t - 1 = 0,\,t + 1 = 0 \\

\Rightarrow t = \dfrac{1}{2},\,t = - 1 \\
$

Now replacing t with the original value, we get –
$
\sin x = \dfrac{1}{2},\,\sin x = - 1 \\

\Rightarrow \sin x = \sin \dfrac{\pi }{6},\,\sin x = - \sin \dfrac{\pi }{2}\,\,\,\,(\sin \dfrac{\pi }{6} =
\dfrac{1}{2},\,\sin \dfrac{\pi }{2} = 1) \\
$

Now,
$
\sin \dfrac{\pi }{6} = \sin (\pi - \dfrac{\pi }{6}) \\

\Rightarrow \sin \dfrac{\pi }{6} = \sin \dfrac{{5\pi }}{6} \\
$

And
$
- \sin \dfrac{\pi }{2} = \sin (\pi + \dfrac{\pi }{2}),\, - \sin \dfrac{\pi }{2} = \sin (2\pi - \dfrac{\pi }{2})
\\

\Rightarrow - \sin \dfrac{\pi }{2} = \sin \dfrac{{3\pi }}{2} \\
$

Thus,
$
\sin x = \sin \dfrac{\pi }{6} = \sin \dfrac{{5\pi }}{6} = \sin \dfrac{{3\pi }}{2} \\

\Rightarrow x = \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} = \dfrac{{3\pi }}{2} \\
$

Hence, the solution of the equation $2{\cos ^2}x - \sin x - 1 = 0$ is $x = \dfrac{\pi }{6}$ or $\dfrac{{5\pi }}{6}$ or $\dfrac{{3\pi }}{2}$ .

Note:The signs of the trigonometric functions are different in different quadrants, we know that the sine function is positive in the first and the second quadrant that’s why we have got 3 answers to the above equation. We are given that the solution lies in the interval $[0,2\pi )$ so the value of x is smaller than $2\pi $ but greater than or equal to zero that’s why we take only these three values as the answer.