First slide

Trigonometric equations

Question

General solution of the equation $\sin^{4} x + \cos^{4} x = \sin x \cos x$ is

Moderate

A

$x = (4 n + 1) \frac{π}{4}$
B

$x = (2 n + 1) \frac{π}{4}$
C

$x = (6 n + 1) \frac{π}{6}$
D

$x = (3 n + 1) \frac{π}{3}$

Solution

We have $\sin^{4} x + \cos^{4} x = \sin x \cos x$
$\Rightarrow 1 - 2 \sin^{2} x \cos^{2} x = \sin x \cos x$
$\Rightarrow 1 - 2 t^{2} = t$ where $t = \sin x \cos x$
$\Rightarrow 2 t^{2} + t - 1 = 0 \Rightarrow 2 t^{2} + 2 t - t - 1 = 0$
$\Rightarrow 2 t (t + 1) - 1 (t + 1) = 0 \Rightarrow (2 t - 1) (t + 1) = 0$
$\Rightarrow t = \frac{1}{2}$ or $t = - 1$
$\begin{array}{l} \Rightarrow \sin x \cos x = \frac{1}{2} o r \sin x \cos x = - 1 \\ \Rightarrow \sin 2 x = 1 o r \sin 2 x = - 2 (n o t p o s s i b l e) \\ \Rightarrow 2 x = 2 n π + \frac{π}{2} \\ \Rightarrow x = n π + \frac{π}{4}, n \in z \\ = (4 n + 1) \frac{π}{4} \end{array}$

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