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Example 1
What is the solution set for the equation $-\cos ^2 x=-2(\sin x+1)$ for $0 < x < 2\pi$ ?
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$\{\pi \}$
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$\{0, \pi \}$
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$\{0, \pi , 2\pi \}$
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$\{0, \dfrac{\pi}{2} , 2\pi \}$
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$\{ \dfrac{3\pi}{2} \}$
Solution
- The above equation may be easily solved if it includes trigonometric of the same type. If $\cos^2 x$ in the given equation is substituted by $1-\sin^2 x$, the given equation will include only $sin x$. Hence our equation becomes
$-(1-\sin ^2 x) = -2(\sin x+1)$
- Expand
$-1+\sin ^2 x = -2\sin x-2$
- Which may be written as
$\sin ^2 x + 2\sin x+1=0$
- With the left side factored as follows
$(\sin x +1)^2=0$
- which gives
$\sin x +1 = 0$
$\sin x = -1$
- The solution set of the above equation for $0 < x < 2\pi$ is
$\{ \dfrac{3\pi}{2} \}$
Answer E
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