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

Hint: 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 \[\left[ -1,1 \right]\].