## Task B-3.4.

If $\frac{3}{1-\sin x-\cos x}=4$, determine what $\cos 4x$ is equal to.
## Solution.

Given the equation $\frac{3}{1-\sin x-\cos x}=4$ after reducing to the form

$\sin x+\cos x=\frac{1}{4}$, we square it. This gives us

$(\sin x+\cos x)^{2}=\frac{1}{16}$,

$\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=\frac{1}{16}$, which simplifies to

$1+2 \sin x \cos x=\frac{1}{16}$. From this, we have

$2 \sin x \cos x=-\frac{15}{16}$, or

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$\sin 2 x=-\frac{15}{16}$. Then,

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$\cos 4 x=1-\sin ^{2} 2 x=1-2 \cdot \frac{225}{256}=-\frac{194}{256}=-\frac{97}{128}$.

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