# Intersection of trigonometric functions

when - 2π < x < 0 .

## 1 Answer

\[2\cos² x= - 5\sin x - 1 \Rightarrow 2(1-\sin^2 x)= - 5\sin x - 1 \]

\[\Rightarrow 2\sin^2 x-5 \sin x-3=0 \Rightarrow (2\sin x +1) (\sin x -3)=0.\]

So $\sin x = -\frac{1}{2}$ or $\sin x=3$. The latter is impossible as $\sin x$ is always a number between $1$ and $-1$. Hence

\[\sin x = -\frac{1}{2} \Rightarrow x=-\frac{\pi}{6}, -\frac{5\pi}{6}.\]

Paul F

191

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