Intersection of trigonometric functions

 Determine the point(s) of intersection of the graphs of y = 2cos²  x and y = - 5sin x - 1
when - 2π < x < 0 .

Trigonometry Functions
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1 Answer

At the intersection points we must have
\[2\cos²  x= - 5\sin x - 1   \Rightarrow    2(1-\sin^2 x)= - 5\sin x - 1  \]
\[\Rightarrow  2\sin^2 x-5 \sin x-3=0   \Rightarrow   (2\sin x +1) (\sin x -3)=0.\]
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 Paul F
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