Number of values of $x$ satisfying $2 {sin}^{2} x + {sin}^{2} 2 x = 2$ and $sin 2 x + cos 2 x = tan x$ , where $x \in [- \frac{3 π}{2}, \frac{4 π}{3}]$ is

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Solution

$2 {sin}^{2} x + {sin}^{2} 2 x = 2 \Rightarrow {sin}^{2} 2 x = 2 {cos}^{2} x \Rightarrow 4 {sin}^{2} x {cos}^{2} x = 2 {cos}^{2} x \Rightarrow {cos}^{2} x = 0 or 2 {sin}^{2} x = 1$
When ${cos}^{2} x = 0$
$\Rightarrow x = {- \frac{3 π}{2}, - \frac{π}{2}, \frac{π}{2}}$
and when ${sin}^{2} x = \frac{1}{2}$
$\Rightarrow x = {\pm \frac{π}{4}, \pm \frac{3 π}{4}, \pm \frac{5 π}{4}}$
$∴ x = {- \frac{3 π}{2}, - \frac{5 π}{4}, - \frac{3 π}{4}, - \frac{π}{2}, - \frac{π}{4}, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{5 π}{4}}$
For
$sin 2 x + cos 2 x = tan x \Rightarrow \frac{2 tan x + 1 - {tan}^{2} x}{1 + {tan}^{2} x} = tan x \Rightarrow {tan}^{3} x + {tan}^{2} x - tan x - 1 = 0 \Rightarrow ({tan}^{2} x - 1) (tan x + 1) = 0 \Rightarrow tan x = - 1, 1 \Rightarrow x = {\pm \frac{π}{4}, \pm \frac{3 π}{4}, \pm \frac{5 π}{4}}$
$∴$ no of common solutions is $6$

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