Solution set of the equation ${sin}^{2} x + {cos}^{2} 3 x = 1$ is given by

{\frac{n π}{4}, n \in I}

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{\frac{n π}{2}, n \in I}

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{n π, n \in I}

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Solution

The correct options are
A ${\frac{n π}{4}, n \in I}$
B ${\frac{n π}{2}, n \in I}$
D ${n π, n \in I}$
${sin}^{2} x + {cos}^{2} 3 x = 1$

Multiply by $2$ on both sides, we get

$2 {sin}^{2} x + 2 {cos}^{2} 3 x = 2$

Using multiple angle formula, $cos 2 x + 1 = 2 {cos}^{2} x$ and $1 - cos 2 x = 2 {sin}^{2} x$ we have

$1 - cos 2 x + cos 6 x + 1 = 2$

$\Rightarrow cos 6 x - cos 2 x = 0$

$\Rightarrow - 2 sin (\frac{- 2 x + 6 x}{2}) sin (\frac{2 x + 6 x}{2}) = 0$

$\Rightarrow - 2 sin 2 x sin 4 x = 0$

$\Rightarrow sin 2 x = 0$ or $sin 4 x = 0$

$\Rightarrow x = n π, 2 x = n π$ or $4 x = n π$

$∴ x = n π$ , $x = \frac{n π}{2}$ , $x = \frac{n π}{4}$

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