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Trigonometry : Trigonometric Equations

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Example Questions

Example Question #26 : Quadratic Formula With Trigonometry

In the interval $0\leq x\leq\pi$ , what values of x satisfy the following equation?

$1-\sin x=\cos^2x-\frac{1}{4}$

Possible Answers:

$x=\frac{\pi}{3},\frac{2\pi}{3}$

$x=\frac{\pi}{4},\frac{7\pi}{12}$

$x=\frac{\pi}{6},\frac{5\pi}{6}$

$x=0,\pi$

Correct answer:

$x=\frac{\pi}{6},\frac{5\pi}{6}$

Explanation:

We start by rewriting the $\cos^2x$ term on the right hand side in terms of $\sin^2x$ .

$\cos^2x=1-\sin^2x$

$1-\sin x=1-\sin^2x-\frac{1}{4}$

We then move everything to the left hand side of the equation and cancel.

$1-\sin x-1+\sin^2x+\frac{1}{4}=0$

$\sin^2x-\sin x+\frac{1}{4}=0$

Apply the quadratic formula:

$\begin{align*} \sin x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(\frac{1}{4})}}{2(1)}\\ &=\frac{1\pm\sqrt{1-1}}{2}\\ &=\frac{1}{2} \end{align*}$

So $\sin x=\frac{1}{2}$ . Using the unit circle, the two values of that yield this are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$ .

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Trigonometry : Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

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Example Question #26 : Quadratic Formula With Trigonometry

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