Question

Solve the following equation for x.
 $\cos 3x{{\cos }^{3}}x+\sin 3x{{\sin }^{3}}x=0$

Answer 1

Hint:
Here we have to find the value of x here. For that, we will find the value of cos3x in terms of cosx and then we will put the value of sin3x in terms of $\sin x$ . Then we will use the distributive law of multiplication to multiply the terms. Then we will break the terms one by one and then we will express the terms in the multiplication of two or more terms and will reduce all the terms and then we will equate the simplified term with zero and find the value of x.

Complete step by step solution:
The given equation is
$\cos 3x{{\cos }^{3}}x+\sin 3x{{\sin }^{3}}x=0$
Now, we will put the value of $\cos 3x$ in terms of and then we will put the value of $\sin 3x$ in terms of $\sin x$.
$\left( 4{{\cos }^{3}}x-3\cos x \right){{\cos }^{3}}x+\left( 3\sin x-4{{\sin }^{3}}x \right){{\sin }^{3}}x=0$
Now, we will use the distributive law of multiplication to multiply the terms.
$4{{\cos }^{6}}x-3{{\cos }^{4}}x+3{{\sin }^{4}}x-4{{\sin }^{6}}x=0$
We can also write the equation as
$3\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)-4\left( {{\sin }^{6}}x-{{\cos }^{6}}x \right)=0$
Now, we will break the terms ${{\cos }^{6}}x-{{\sin }^{6}}x\And {{\cos }^{4}}x+{{\sin }^{4}}x$further.
$3\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)-4\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x \right)=0$
We will put the value of$\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$and simplify the terms further.
$3\left( -\cos 2x \right)\left( 1 \right)-4\left( -\cos 2x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x \right)=0$
Taking cos2x common, we get
$\cos 2x\left[ -3+4\left( {{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x \right) \right]=0$
We can also write the equation as
 $\cos 2x\left[ -3+4\left( {{\sin }^{2}}x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+{{\cos }^{4}}x \right) \right]=0$
We will put the value of$\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$now.
$\cos 2x\left[ -3+4\left( {{\sin }^{2}}x+{{\cos }^{4}}x \right) \right]=0$
We know, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Therefore,
$\begin{align}
  & \cos 2x\left[ -3+4\left( 1-{{\cos }^{2}}x+{{\cos }^{4}}x \right) \right]=0 \\
 & \cos 2x\left( 4{{\cos }^{4}}x-4{{\cos }^{2}}x+1 \right)=0 \\
 & \cos 2x{{\left( 2{{\cos }^{2}}x-1 \right)}^{2}}=0 \\
\end{align}$
We know, $2{{\cos }^{2}}x-1=\cos 2x$. Therefore,
${{\cos }^{3}}2x=0$
Thus,
$\cos 2x=0$
The solution of cos2x is
$\begin{align}
  & 2x=\dfrac{\pi }{2}+2n\pi \\
 & x=\dfrac{\pi }{4}+n\pi \\
\end{align}$
This is the requirement of x.

Note:
Since we have used the distributive law of multiplication. Let’s define it to understand its meaning.
Let a, b and c be three numbers, then according to distributive law of multiplication;
$a(b+c)=a.b+a.c$
We have calculated the value of x by the solving the equation $\cos 2x=0$
The general solution of $\cos x=0$is $x=\dfrac{\pi }{2}+2n\pi $. Here we have replaced x with 2x and solved the equation and got the value of x.