Solve the following equations.
$s i n^{2} x + s i n^{2} 2 x + s i n^{2} 3 x + s i n^{2} 4 x = 2.$

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Solution

${sin}^{2} x + {sin}^{2} 2 x + {sin}^{2} 3 x + {sin}^{2} 4 x = 2$
Multiply $2$ on both sides of the equation and use the identity $cos 2 x = 1 - 2 {sin}^{2} x$
$=> 2 {sin}^{2} x + 2 {sin}^{2} 2 x + 2 {sin}^{2} 3 x + 2 {sin}^{2} 4 x = 4 => 1 - cos 2 x + 1 - cos 4 x + 1 - cos 6 x + 1 - cos 8 x = 4 => 4 - (cos 2 x + cos 4 x + cos 6 x + cos 8 x) = 4 => cos 2 x + cos 4 x + cos 6 x + cos 8 x = 0 => cos 2 x + cos 8 x + cos 4 x + cos 6 x = 0 => 2 cos 5 x cos 3 x + 2 cos 5 x cos x = 0 => 2 cos 5 x (cos 3 x + cos x) = 0 => 4 cos 5 x \times cos 2 x \times cos x = 0 cos 5 x = 0 cos 2 x = 0 cos x = 0 => 5 x = \frac{(2 n + 1) π}{2} => 2 x = \frac{(2 m + 1) π}{2} => x = \frac{(2 k + 1) π}{2} => x = \frac{(2 n + 1) π}{10} => x = \frac{(2 m + 1) π}{4}$
$n, m, k$ belongs to integers.

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