Trigonometrical Equations Ques 16
- The number of distinct solutions of the equation $\frac{5}{4} \cos ^{2} 2 x+\cos ^{4} x+\sin ^{4} x+\cos ^{6} x+\sin ^{6} x=2$ in the interval $[0,2 \pi]$ is
(2015 Adv.)
Show Answer
Answer:
Correct Answer: 16.(8)
Solution:
Formula:
- Here, $\frac{5}{4} \cos ^{2} 2 x+\left(\cos ^{4} x+\sin ^{4} x\right)+\left(\cos ^{6} x+\sin ^{6} x\right)=2$
$\Rightarrow \quad \frac{5}{4} \cot 2 x+\left[\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right]$
$+\left(\cos ^{2} x+\sin ^{2} x\right)\left[\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x\right]=2$
$\Rightarrow \frac{5}{4} \cos ^{2} 2 x+\left(1-2 \sin ^{2} x \cos ^{2} x\right)+\left(1-3 \cos ^{2} x \sin ^{2} x\right)=2$
$\Rightarrow \quad \frac{5}{4} \cos ^{2} 2 x-5 \sin ^{2} x \cos ^{2} x=0$
$ \begin{array}{lc} \Rightarrow & \frac{5}{4} \cos ^{2} 2 x-\frac{5}{4} \sin ^{2} 2 x=0 \\ \Rightarrow & \frac{5}{4} \cos ^{2} 2 x-\frac{5}{4}+\frac{5}{4} \cos ^{2} 2 x=0 \\ \Rightarrow & \frac{5}{2} \cos ^{2} 2 x=\frac{5}{4} \Rightarrow \cos ^{2} 2 x=\frac{1}{2} \\ \Rightarrow & 2 \cos ^{2} 2 x=1 \\ \Rightarrow & \quad 1+\cos 4 x=1 \\ \Rightarrow & \cos 4 x=0 \text { as } 0 \leq x \leq 2 \pi \\ \therefore & 4 x= \{ \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2}, \frac{13 \pi}{2}, \frac{15 \pi}{2} \} \\ \text { as } & 0 \leq 4 x \leq 8 \pi \\ \Rightarrow & x= \{ \frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}, \frac{15 \pi}{8} \} \end{array} $
Hence, the total number of solutions is 8 .
BETA
