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 .
when - 2π < x < 0 .
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
\[\sin x = -\frac{1}{2} \Rightarrow x=-\frac{\pi}{6}, -\frac{5\pi}{6}.\]
\[2\cos² x= - 5\sin x - 1 \Rightarrow 2(1-\sin^2 x)= - 5\sin x - 1 \]
\[\Rightarrow 2\sin^2 x-5 \sin x-3=0 \Rightarrow (2\sin x +1) (\sin x -3)=0.\]
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
200
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