Question

How do you solve $2{{\sin }^{2}}x=1+\cos x$ for ${{0}^{\circ }}\le x\le {{180}^{\circ }}$?

Answer 1

To solve the given trigonometric equation, we first need to convert ${{\sin }^{2}}x$ in terms of $\cos x$ using the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Using this, we will obtain a trigonometric equation in terms of $\cos x$. Then we need to substitute $\cos x=t$ to get a quadratic equation in $t$. On solving the quadratic equation, we will get two values of $t$, which will correspond to the values of $\cos x$. Then using the given range, we can determine the final solution of the equation.

Complete step by step answer:
The equation given in the question is
$2{{\sin }^{2}}x=1+\cos x............(i)$
As we can observe in the above equation that it contains two trigonometric functions, ${{\sin }^{2}}x$ and $\cos x$. For solving a trigonometric equation, we first have to convert it in the form of only one trigonometric function. Now, clearly $\cos x$ can’t be converted in any form so it will remain as it is. But we can convert ${{\sin }^{2}}x$ in terms of $\cos x$ by using the identity
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Subtracting ${{\cos }^{2}}x$ from both the sides, we get
$\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x.............(ii)$
Putting equation (ii) in (i), we get

& \Rightarrow 2\left( 1-{{\cos }^{2}}x \right)=1+\cos x \\\ & \Rightarrow 2-2{{\cos }^{2}}x=1+\cos x \\\ \end{aligned}$$ Adding $2{{\cos }^{2}}x-2$ both the sides, we get $$\begin{aligned} & \Rightarrow 2-2{{\cos }^{2}}x+2{{\cos }^{2}}x-2=1+\cos x+2{{\cos }^{2}}x-2 \\\ & \Rightarrow 0=2{{\cos }^{2}}x+\cos x-1 \\\ & \Rightarrow 2{{\cos }^{2}}x+\cos x-1=0..........(iii) \\\ \end{aligned}$$ Substituting $\cos x=t$ in the above equation, we get $$\Rightarrow 2{{t}^{2}}+t-1=0$$ So we have a quadratic equation in $t$. Now, the above equation can be written as $$\Rightarrow 2{{t}^{2}}+2t-t-1=0$$ Taking $2t$ common from the first two terms, and $-1$ common from the last two terms, we get $$\Rightarrow 2t\left( t+1 \right)-1\left( t+1 \right)=0$$ Now taking $$\left( t+1 \right)$$ common, we have $$\begin{aligned} & \Rightarrow \left( t+1 \right)\left( 2t-1 \right)=0 \\\ & \Rightarrow \left( t+1 \right)=0,\left( 2t-1 \right)=0 \\\ \end{aligned}$$ On solving we get $\Rightarrow t=-1,t=\dfrac{1}{2}$ Now, according to our substitution, $t=\cos x$. This means that $\cos x=-1,\cos x=\dfrac{1}{2}$ According to the question, the interval of $x$ is given to be $${{0}^{\circ }}\le x\le {{180}^{\circ }}$$. This means that $x$ can take the values from the first two quadrants. We know that $\cos x$ is positive in the first quadrant, and is negative in the second quadrant. So the negative value $-1$ must belong to the second quadrant, and the positive value $\dfrac{1}{2}$ must belong to the first quadrant. From the first part of the solution, we have $\begin{aligned} & \Rightarrow \cos x=-1 \\\ & \Rightarrow x={{180}^{\circ }} \\\ \end{aligned}$ And from the second part of the solution we have $\begin{aligned} & \Rightarrow \cos x=\dfrac{1}{2} \\\ & \Rightarrow x={{60}^{\circ }} \\\ \end{aligned}$ So the solution of the given equation $$2{{\sin }^{2}}x=1+\cos x$$ is $x={{180}^{\circ }}$ and $x={{60}^{\circ }}$. **Note:** Do not end your solution after writing the values for $\cos x$.The variable of the equation is $x$, not $\cos x$. So the solution of the equation means the values of the variable $x$. Also, be careful to make sure that the solutions to the given trigonometric equation obtained must be in the range specified in the question.