QJEEAdv15Ma87a3: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma87a3 Solution: The number of distinct solutions of the equation in the…

Given: $\frac{5}{4} \cos ^2 2 x+\cos ^4 x+\sin ^4 x+\cos ^6 x+$ $\sin ^6 x=2$

$\begin{aligned}\Rightarrow & \frac{5}{4} \cos ^2 2 x+\left(\cos ^2 x\right)^2+\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^3 \\\\& +\left(\sin ^2 x\right)^3=2\quad\quad...(i)\end{aligned}$

As we know, $a^2+b^2+2 a b=(a+b)^2$

$\Rightarrow a^2+b^2=(a+b)^2-2 a b$

And $a^3+b^3+3 a b(a+b)=(a+b)^3$

$\Rightarrow a^3+b^3=(a+b)^3-3 a b(a+b)$

So, equation (i) can be written as

$\begin{aligned}& \frac{5}{4} \cos ^2 2 x+\left(\cos ^2 x+\sin ^2 x\right)^2-2\left(\cos ^2 x\right) \\\\&\left(\sin ^2 x\right)+\left(\cos ^2 x+\sin ^2 x\right)^3-3 \cos ^2 x \sin ^2 x \\\\&\left(\cos ^2 x+\sin ^2 x\right)=2 \\\\& \Rightarrow \frac{5}{4} \cos ^2 2 x+(1)^2-2 \cos ^2 x \sin ^2 x+(1)^3 \\\\&-3 \cos ^2 x \sin ^2 x(1)=2~~~~\left\{\because \cos ^2 x+\sin ^2 x=1\right\}\end{aligned}$

$\begin{aligned}& \Rightarrow \frac{5}{4} \cos ^2 2 x+2-5 \cos ^2 x \sin ^2 x=2 \\\\& \Rightarrow \frac{5}{4} \cos ^2 2 x-5 \cos ^2 x \sin ^2 x=0\end{aligned}$

As we know, $\sin 2 \theta=2 \sin \theta \cos \theta$

$\begin{aligned}& \Rightarrow \frac{5}{4} \cos ^2 2 x-\frac{5}{4} \sin ^2 2 x=0 \\\\& \because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta \\\\& \Rightarrow \frac{5}{4} \cos 4 x=0 \\\\& \Rightarrow \cos 4 x=0 \\\\& \Rightarrow 4 x=2(n+1) \frac{\pi}{2}, n \in \mathrm{I} \\\\& \because x \in[0,2 \pi]\end{aligned}$

So, possible distinct values of $x$ are $\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}$ and $\frac{15 \pi}{8}$ .

So, the number of distinct solutions of the given equation are 8 .