Question

How do you solve $2{{\cos }^{2}}x+\sin x-2=0$ over the interval $0$ to $2\pi $.

Answer 1

Hint: In the given equation we have one ${{\cos }^{2}}$ term and one $\sin $ term. So, we need to convert the whole equation in terms of $\cos x$ or $\sin x$. For this we are going to use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. From this identity we will substitute the value of ${{\cos }^{2}}x$ as $1-{{\sin }^{2}}x$ in the given equation and convert the given equation in form of $a{{x}^{2}}+bx+c=0$ . Now we will simplify and factorize the given equation, to get the solution of the given equation.

Complete step by step answer:
Given that, $2{{\cos }^{2}}x+\sin x-2=0$.
We have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. From this identity the value of ${{\cos }^{2}}x$ is $1-{{\sin }^{2}}x$. Substituting this value in the given equation, then we will get
$2\left( 1-{{\sin }^{2}}x \right)+\sin x-2=0$
Simplifying the above equation, then we will get
$\begin{align}
  & 2-2{{\sin }^{2}}x+\sin x-2=0 \\
 & \Rightarrow -2{{\sin }^{2}}x+\sin x=0 \\
\end{align}$
Multiplying the above equation with negative sign, then we will get
$2{{\sin }^{2}}x-\sin x=0$
Taking $\sin x$ common from the above equation, then we will get
$\Rightarrow \sin x\left( 2\sin x-1 \right)=0$
Equating each term to zero individually, then we will have
$\sin x=0$ or $2\sin x-1=0\Rightarrow \sin x=\dfrac{1}{2}$
In the problem they mentioned the interval as $\left[ 0,2\pi \right]$.
So, in the given interval we will have $\sin x=0$ for $x=\left\{ 0,\pi ,2\pi \right\}$.
In the given interval we will have $\sin x=\dfrac{1}{2}$ for $x=\left\{ \dfrac{\pi }{6},\dfrac{5\pi }{6} \right\}$

So, the set of solutions for the equation $2{{\cos }^{2}}x+\sin x-2=0$ in the interval $\left[ 0,2\pi \right]$ is $\left\{ 0,\dfrac{\pi }{6},\dfrac{5\pi }{6}\pi ,2\pi \right\}$.

Note: For this problem we can also convert $\sin x$ into ${{\cos }^{2}}x$ and solve the problem. But there you will get a complex equation and the solution is not obtained easily and also there are a lot of chances to make mistakes. So, we have not chosen that method. You can also draw a graph for the given equation and check the points where the given equation meets the $x-axis$ and those points are solutions for the equation.