Question

How do you solve $1-\sin 2x=\cos x-\sin x$?

Answer 1

To solve the given question, we should know some of the trigonometric properties/ formulas. The first is the trigonometric identity which states that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We should also know the trigonometric formula $\sin 2x=2\sin x\cos x$. We should also know the general solution for $\sin x=a$, the general solution for this is ${{\left( -1 \right)}^{k}}\theta +\pi k$here $\theta$ is a standard solution in the range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.

Complete step by step answer:
We are given the equation $1-\sin 2x=\cos x-\sin x$, we need to find the values of x that satisfies the equation,
Squaring both sides of the above equation, we get
$\Rightarrow {{\left( 1-\sin 2x \right)}^{2}}={{\left( \cos x-\sin x \right)}^{2}}$
Expanding the brackets on both sides, we get
$\Rightarrow 1+{{\sin }^{2}}\left( 2x \right)-2\sin \left( 2x \right)={{\cos }^{2}}x+{{\sin }^{2}}x-2\cos x\sin x$
As we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, and $\sin 2x=2\sin x\cos x$. Using these the above equation can be expressed as
$\Rightarrow 1+{{\sin }^{2}}\left( 2x \right)-2\sin \left( 2x \right)=1-\sin \left( 2x \right)$
Taking all the terms of the equation to the left side of the equation, we get
$\Rightarrow 1+{{\sin }^{2}}\left( 2x \right)-2\sin \left( 2x \right)-1+\sin \left( 2x \right)=0$
Simplifying the above equation, we get
$\Rightarrow {{\sin }^{2}}\left( 2x \right)-\sin \left( 2x \right)=0$
Substituting $\sin \left( 2x \right)=t$ in the above equation, it can be expressed as
$\Rightarrow {{t}^{2}}-t=0$
As we can see that this is a quadratic equation in the variable $t$. we need to find the values of $t$ that satisfy the equation. The constant term of the equation is zero, so one root of the equation must be zero. Using this we can express the quadratic in factored form as
$\Rightarrow \left( t-0 \right)\left( {{t}^{{}}}-1 \right)=0$
Hence the roots of the equation are $t=0$ or ${{t}^{{}}}=1$. Using the substitution $\sin \left( 2x \right)=t$, we get $\sin 2x=0$ or $\sin 2x=1$. Using the general solution of $\sin x=a$, that is ${{\left( -1 \right)}^{k}}\theta +\pi k$here $\theta$ is a standard solution in the range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
The standard solution of $\sin 2x=0$ is $0$, and $\sin 2x=1$ is $\dfrac{\pi }{2}$.
For $\sin 2x=0$, $2x={{\left( -1 \right)}^{k}}0+\pi k$
$\Rightarrow x=\dfrac{\pi k}{2}$, $k$ is an integer
For $\sin 2x=1$, $2x={{\left( -1 \right)}^{k}}\dfrac{\pi }{2}+\pi k$
$\Rightarrow x={{\left( -1 \right)}^{k}}\dfrac{\pi }{4}+\dfrac{\pi k}{2}$, $k$ is an integer.

Note:
The trigonometric identities and properties of all trigonometric functions should be remembered to solve these types of questions. Rather than a general solution, we can also find all the solutions present in the range of $\left[ 0,2\pi \right]$.