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)

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Answer:

Correct Answer: 2. (b)

Solution:

  1. 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.



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