Trigonometrical Equations 1 Question 2
2. The number of solutions of the equation $1+\sin ^{4} x=\cos ^{2} 3 x, x \in-\frac{5 \pi}{2}, \frac{5 \pi}{2}$ is
(a) 3
(b) 5
(c) 7
(d) 4
(2019 Main, 12 April I)
Show Answer
Answer:
Correct Answer: 2. (b)
Solution:
- Given equation is $1+\sin ^{4} x=\cos ^{2}(3 x)$
Since, range of $\left(1+\sin ^{4} x\right)=[1,2]$
and range of $\cos ^{2}(3 x)=[0,1]$
So, the given equation holds if
$$ \begin{array}{ll} & 1+\sin ^{4} x=1=\cos ^{2}(3 x) \\ \Rightarrow \quad & \sin ^{4} x=0 \text { and } \cos ^{2} 3 x=1 \end{array} $$
Since, $x \in-\frac{5 \pi}{2}, \frac{5 \pi}{2}$
$$ \therefore \quad x=-2 \pi,-\pi, 0, \pi, 2 \pi . $$
Thus, there are five different values of $x$ is possible.
