Question

Solve the following equations.
${{\sin }^{2}}x+{{\sin }^{2}}2x-{{\sin }^{2}}3x-{{\sin }^{2}}4x=0$

Answer 1

First of all rearrange the equation as $\left( {{\sin }^{2}}x-{{\sin }^{2}}3x \right)+\left( {{\sin }^{2}}2x-{{\sin }^{2}}4x \right)=0$. Now use the formula ${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right)\sin \left( A-B \right)$ to simplify the given equation. Now, equate the factors to 0 and use $\sin \theta =\sin \alpha$, then $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$ and $\cos \theta =\cos \alpha$, then $\theta =2n\pi \pm \alpha$ to find the solution of the equations.

Complete step-by-step answer:
In this question, we have to solve the equation,
${{\sin }^{2}}x+{{\sin }^{2}}2x-{{\sin }^{2}}3x-{{\sin }^{2}}4x=0$
Let us consider the equation given in the question.
${{\sin }^{2}}x+{{\sin }^{2}}2x-{{\sin }^{2}}3x-{{\sin }^{2}}4x=0$
By rearranging the equation, we get,
$\left( {{\sin }^{2}}x-{{\sin }^{2}}3x \right)+\left( {{\sin }^{2}}2x-{{\sin }^{2}}4x \right)=0$
We know that
${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right)\sin \left( A-B \right)$
By using this in the above equation, we get,
$\sin \left( x+3x \right)\sin \left( x-3x \right)+\sin \left( 2x+4x \right)\sin \left( 2x-4x \right)=0$
By simplifying the above equation, we get,
$\sin 4x\sin \left( -2x \right)+\sin 6x\sin \left( -2x \right)=0$
By taking out sin (– 2x) common we get,
$\sin \left( -2x \right)\left[ \sin 4x+\sin 6x \right]=0$
We know that,
$\sin \left( -\theta \right)=-\sin \theta$
By using this, we get,
$\left( -\sin 2x \right)\left[ \sin 4x+\sin 6x \right]=0$
From this, we get,
$\sin 2x=0;\left( \sin 4x+\sin 6x \right)=0$
Let us take, sin 2x = 0
We know that sin 0 = 0. So, we get,
$\sin 2x=\sin {{0}^{o}}$
We know that when $\sin \theta =\sin \alpha$, then $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$
So, we get,
$2x=n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)$
$2x=n\pi$
So, $x=\dfrac{n\pi }{2}\text{ where }n\in I...\left( i \right)$
Now, let us take sin 4x + sin 6x = 0
We know that,
$\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
By using this, we get,
$2\sin \left( \dfrac{4x+6x}{2} \right)\cos \left( \dfrac{4x-6x}{2} \right)=0$
So, $\sin \left( \dfrac{10x}{2} \right)\cos \left( \dfrac{-2x}{2} \right)=0$
$\sin 5x\cos \left( -x \right)=0$
We know that $\cos \left( -\theta \right)=\cos \theta$. By using this, we get,
$\sin 5x\cos x=0$
From this we get, sin 5x = 0 and cos x = 0
Let us take sin 5x = 0
We know that sin 0 = 0. From this, we get,
sin 5x = sin 0
We know that when $\sin \theta =\sin \alpha$, then $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$. So, we get,
$5x=n\pi +{{\left( -1 \right)}^{n}}0$
$5x=n\pi$
So, $x=\dfrac{n\pi }{5}\text{ where }n\in I....\left( ii \right)$
Let us take cos x = 0. We know that $\cos \dfrac{\pi }{2}=0$. From this, we get,
$\cos x=\cos \dfrac{\pi }{2}$
We know that when $\cos \theta =\cos \alpha$, then $\theta =2n\pi \pm \alpha$. So, we get,
$x=n\pi \pm \dfrac{\pi }{2}\text{ where }n\in I....\left( iii \right)$
So, from equation (i), (ii) and (iii), we get,
$x\in \left( n\dfrac{\pi }{2} \right)\cup \left( \dfrac{n\pi }{5} \right)\cup \left( n\pi \pm \dfrac{\pi }{2} \right)\text{where }n\in I$

Note: Students are advised to memorize the formulas in trigonometry as they come handy while solving questions which saves a lot of time and lengthy calculations. Also, at the end of this question, some students make this mistake of taking the intersection of the three solutions instead of union which is wrong because each one of these solutions will satisfy the given equations and not just the solutions which are intersections of three solutions or common to these solutions. In this question, students can check their solution by taking some integral value of n and satisfying it in the given equation.