Question

How do you solve $\sin \left( \dfrac{x}{2} \right)+\cos x=1$?

Answer 1

Hint : In the question we have two trigonometric ratios first one is $\sin$ and the second one is $\cos$. To solve the given equation, we need to convert the whole equation in terms of a single trigonometric ratio i.e., we need to convert the terms in the equations into either $\sin$ or $\cos$. For this we will subtract the term $\cos x$ on both sides of the equation, then we will get $1-\cos x$ in LHS. We have the formula $1-\cos x=2{{\sin }^{2}}\left( \dfrac{x}{2} \right)$. So, we will substitute this value in the given equation and solve the obtained equation to get the required result.

Complete step by step answer:
Given that, $\sin \left( \dfrac{x}{2} \right)+\cos x=1$.
Subtracting $\cos x$ from both sides of the above equation, then we will get
$\sin \left( \dfrac{x}{2} \right)+\cos x-\cos x=1-\cos x$
We know that $+x-x=0$, then we will have
$\Rightarrow \sin \left( \dfrac{x}{2} \right)=1-\cos x$
We have a trigonometric formula $1-\cos x=2{{\sin }^{2}}\left( \dfrac{x}{2} \right)$. Substituting this value in the above equation, then we will get
$\Rightarrow \sin \left( \dfrac{x}{2} \right)=2{{\sin }^{2}}\left( \dfrac{x}{2} \right)$
Simplifying the above equation, then we will get
$\Rightarrow \sin \left( \dfrac{x}{2} \right)-2{{\sin }^{2}}\left( \dfrac{x}{2} \right)=0$
Taking $\sin \left( \dfrac{x}{2} \right)$ as common in LHS of the above equation, then we will get
$\Rightarrow \sin \left( \dfrac{x}{2} \right)\left[ 1-2\sin \left( \dfrac{x}{2} \right) \right]=0$
Equating each term to zero individually, then we will have
$\sin \left( \dfrac{x}{2} \right)=0$ or $\begin{aligned} & 1-2\sin \left( \dfrac{x}{2} \right)=0 \\\ & \Rightarrow 2\sin \left( \dfrac{x}{2} \right)=1 \\\ & \Rightarrow \sin \left( \dfrac{x}{2} \right)=\dfrac{1}{2} \\\ \end{aligned}$
We have the values $\sin 0=\sin \pi =\sin 2\pi =0$, $\sin \dfrac{\pi }{6}=\sin \dfrac{5\pi }{6}=\dfrac{1}{2}$. So the equation $\sin \left( \dfrac{x}{2} \right)=0$ has three solution which are
$\dfrac{x}{2}=0\Rightarrow x=0$
$\dfrac{x}{2}=\pi \Rightarrow x=2\pi$
$\dfrac{x}{2}=2\pi \Rightarrow x=4\pi$.
Now the equation $\sin \left( \dfrac{x}{2} \right)$ has two solutions which are
$\dfrac{x}{2}=\dfrac{\pi }{6}\Rightarrow x=\dfrac{\pi }{3}$
$\dfrac{x}{2}=\dfrac{5\pi }{6}\Rightarrow x=\dfrac{5\pi }{3}$
Hence the solution set for the given equation $\sin \left( \dfrac{x}{2} \right)+\cos x=1$ is $x=0,\dfrac{\pi }{3},\dfrac{5\pi }{3},2\pi ,4\pi$.

Note:
We can also solve the trigonometric equations by plotting a graph on the coordinate system. We can observe that the graph is looks like below

The solutions for the given equation are the points where the plot meets the $x-axis$.