Question

Find the number of solutions of the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ where $x\in \left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$.
(a) 5
(b) 4
(c) 7
(d) 3

Answer 1

In this question, we are given with the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$. Now for in order to find the number of solutions of the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ where $x\in \left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$
We should know the following properties of the trigonometric function, that is the range of the trigonometric functions $\sin x$ and $\cos x$. We have that the values of $\sin x$ lie in the interval $\left[ -1,1 \right]$ and the values of $\cos x$ also lie in the interval $\left[ -1,1 \right]$. Using this we will have to find the common values of $x$ in the interval $\left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$ such that the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ holds true.

Complete step by step answer:
We are given with the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$.
Now we know that the value of $\sin x$ lies in the interval $\left[ -1,1 \right]$.
That is, we have
$-1\le \sin x\le 1$
Now since the values of ${{\sin }^{2}}x$ is always positive, then we must have
$0\le {{\sin }^{2}}x\le 1$
Therefore we can also have
$0\le {{\sin }^{4}}x\le 1$
In order to find the range of values of the function $1+{{\sin }^{4}}x$, we will add 1 on all the terms in the inequality $0\le {{\sin }^{4}}x\le 1$ to get
$1\le 1+{{\sin }^{4}}x\le 2$
Therefore we have
$1\le 1+{{\sin }^{4}}x.........(1)$
Now we know that the value of $\cos x$ lies in the interval $\left[ -1,1 \right]$.
That is, we have
$-1\le \cos x\le 1$
Then we also have
$-1\le \cos 3x\le 1$
Now since the values of ${{\cos }^{2}}3x$ is always positive, then we must have
$0\le {{\cos }^{2}}3x\le 1$
Therefore we can also have
${{\cos }^{2}}3x\le 1............(2)$
Using inequality (1) and inequality (2), we have that that the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ holds only when we have
$1+{{\sin }^{4}}x=1$ and ${{\cos }^{2}}3x=1$.
Now if we have $1+{{\sin }^{4}}x=1$, then this implies that
${{\sin }^{4}}x=0$
Since we know that ${{x}^{k}}=0$ implies $x=0$ for any natural number $k$.
Therefore we have $\sin x=0$.
Now since we also know that the value of $\sin x=0$ if and only if $x=n\pi$ for any integer value of $n$.
Therefore for $x\in \left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$, we can have that $\sin x=0$ if and only if $x$ takes values $-2\pi ,-\pi ,0,\pi ,2\pi$.
Now we also know that $\cos n\pi =1$ if and only if $x=n\pi$.
Thus ${{\cos }^{2}}3n\pi$ is also equal to 1 since $3n$ is also an integer.
Therefore for all value of $x$ given by $-2\pi ,-\pi ,0,\pi ,2\pi$, we have
${{\cos }^{2}}3x=1$
Hence the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ is satisfied for five values of $x$ given by $-2\pi ,-\pi ,0,\pi ,2\pi$ in the interval $\left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$.
That is the number of solutions of the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ where $x\in \left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$ is equals to 5.

So, the correct answer is “Option A”.

Note: In this problem, in order to determine the number of solutions of the equation $1+{{\sin }^{4}}x={{\cos }^{2}}3x$ where $x\in \left[ -\dfrac{5\pi }{2},\dfrac{5\pi }{2} \right]$ we have to check the values of $x$ for which the trigonometric functions $1+{{\sin }^{4}}x$ and ${{\cos }^{2}}3x$ are equal. Also we should know that the values of $\sin x$ lies in the interval $\left[ -1,1 \right]$ and the values of $\cos x$ also lie in the interval $\left[ -1,1 \right]$.