Question

How do you find all the solutions for the trigonometric equation $\sin 2x = \cos x$ in the interval $[0,2\pi ]$ ?

Answer 1

Hint: The given question involves solving a trigonometric equation and finding values of angle x that satisfy the given equation and lie in the range of $[0,2\pi ]$. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities. We will make use of basic trigonometric formulae $\sin 2x = 2\sin x\cos x$ to solve the trigonometric equation.

Complete answer:
In the given problem, we have to solve the trigonometric equation $\sin 2x = \cos x$ and find the values of x that satisfy the given equation and lie in the range of $[0,2\pi ]$.
So, in order to solve the given trigonometric equation $\sin 2x = \cos x$, we should first take all the terms to the left side of the equation.
Transposing all the terms to left side of the equation, we get,
$ \Rightarrow \sin 2x - \cos x = 0$
Now, we know the double angle formula for sine, $\sin 2x = 2\sin x\cos x$ . Hence, substituting $2\sin x\cos x$ as $\sin \left( {2x} \right)$, we get,
\[ \Rightarrow 2\sin x\cos x - \cos x = 0\]
Taking cosine common from both the terms, we get,
\[ \Rightarrow \cos x\left( {2\sin x - 1} \right) = 0\]
Now, for the product of two terms to be equal to zero, either one or both the terms have to be zero.
Hence, either \[\cos x = 0\] or \[\left( {2\sin x - 1} \right) = 0\]
Shifting the terms, we get,
Either \[\cos x = 0\] or \[\sin x = \dfrac{1}{2}\]
Now, we know that the cosine function is zero for the odd multiples of $\left( {\dfrac{\pi }{2}} \right)$. So, we get, $x = \left( {2n + 1} \right)\dfrac{\pi }{2}$ for \[\cos x = 0\].
Now, we need to find values of x in the range $[0,2\pi ]$. So, substituting the values of n as $0$ and $1$, we get,
$x = \left( {2 \times 0 + 1} \right)\dfrac{\pi }{2}$ and $x = \left( {2 \times 1 + 1} \right)\dfrac{\pi }{2}$
So, we get, $x = \dfrac{\pi }{2}$ and $x = \dfrac{{3\pi }}{2}$.
Now, we solve for x in \[\sin x = \dfrac{1}{2}\].
We know that the value of $\sin \dfrac{\pi }{6}$ is \[\dfrac{1}{2}\]. We also know that the sine trigonometric function is positive in the first and second quadrants. Also, $\sin \left( {\pi - x} \right) = \sin x$.
So, we have the values of x satisfying the equation \[\sin x = \dfrac{1}{2}\] and lying in the range $[0,2\pi ]$ as: $\dfrac{\pi }{6}$ and $\left( {\pi - \dfrac{\pi }{6}} \right) = \dfrac{{5\pi }}{6}$.
Therefore, all the solutions of the trigonometric equation $\sin 2x = \cos x$ in the interval $[0,2\pi ]$ are: $\dfrac{\pi }{6}$, $\dfrac{\pi }{2}$, $\dfrac{{5\pi }}{6}$ and $\dfrac{{3\pi }}{2}$.

Note:
We should know the formula for finding the general solutions for $\sin x = \sin y$, $\cos x = \cos y$ and $\tan x = \tan y$ for solving such questions. Then, we substitute the value of the parameter in the general solutions of the equations to find the principal solution lying in the range of $[0,2\pi ]$. We should have a clear understanding of trigonometric formulae, identities and simplification rules to tackle such questions.