Question

How do you solve $2{\sin ^2}x = 2 + \cos x$ between $0$ and $2pi$?

Answer 1

In this problem we have given an equation and we asked to solve the given equation between the limits $0$ and $2pi$. There is a sine term in the equation and to solve this given equation we need to replace the sine term by $\cos$ term. Because in the right hand side there is a $\cos$ term so making the common terms on both sides makes us solve the problem easily.

Formula used: ${\sin ^2}x = 1 - {\cos ^2}x$

Complete step-by-step solution:
Given equation is $2{\sin ^2}x = 2 + \cos x$ between $0$ and $2pi$
First we are going to substitute ${\sin ^2}x = 1 - {\cos ^2}x$ in the given equation to remove the sine term from the left hand side of the given equation, we get
$\Rightarrow 2\left( {1 - {{\cos }^2}x} \right) = 2 + \cos x$, multiply $2$ with each term in the left hand side, implies
$\Rightarrow 2 - 2{\cos ^2}x = 2 + \cos x$, now keep all the cosine terms in the right hand side and take the integers to left hand side, we get
$\Rightarrow 2 - 2 = \cos x + 2{\cos ^2}x$, here $+ 2$ and $- 2$ gets cancelled and so the left hand side becomes $0$.
$\Rightarrow 0 = \cos x + 2{\cos ^2}x - - - - - \left( 1 \right)$
In the right hand side of equation (1) $\cos x$ is common in both the addition of two terms. So we are going to take one $\cos x$ as common, we get
$\Rightarrow 0 = \cos x\left( {1 + 2\cos x} \right)$
Now, we are going to zero product property, then
$\Rightarrow \cos x = 0$ or $2\cos x + 1 = 0$
$\Rightarrow \cos x = 0$ or $2\cos x = - 1$
$\Rightarrow \cos x = 0$ or $\cos x = \dfrac{{ - 1}}{2}$, here our work is to find the value of $x$.
Now take the $\cos x$ to the right hand side, then it becomes $\cos x$ of inverse, that is ${\cos ^{ - 1}}$, we get
$\Rightarrow x = {\cos ^{ - 1}}\left( 0 \right)$Or$x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

$\Rightarrow x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$Or$x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}$
This is the required answer.

Note: Here ${\cos ^{ - 1}}$ is assuming that the inverse of cosine. And ${\cos ^{ - 1}}$ is an angle $x$, for which $\cos x = 0$. If we take the graph of $f\left( x \right) = \cos x$ if we see that $\cos x = 0$ then $x = \dfrac{\pi }{2} + k\pi$ for any $k \in Z$, so that the smallest positive value is $\dfrac{\pi }{2}$ or $\dfrac{{3\pi }}{2}$. Like the same way we use for $x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ and we get $x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}$.