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Trigonometry

Lesson 2: Sinusoidal equations

Solve sinusoidal equations (basic)

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Jamie tried to solve the following equation, where

x

is in radians to the nearest hundredth, in the interval from

- π

π

, but she made a mistake.

\cos (x) = - 0.08

Here is Jamie's work.

\begin{array}{rcl} \cos^{- 1} (\cos (x)) & = & \cos^{- 1} (- 0.08) \\ x & \approx & 1.65 \\ ↙ & ↘ \\ \begin{aligned} Using \cos (x) = \cos (x + 2 π) \\ x & \approx 1.65 + 2 π n \\ x & \approx \dots - 4.63, 1.65, 7.93 \dots \end{aligned} & \begin{aligned} Using \cos (x) = \cos (π - x) \\ and \cos (x) = \cos (x + 2 π) \\ x & \approx 1.49 + 2 π n \\ x & \approx \dots - 4.79, 1.49, 7.77 \dots \end{aligned} \end{array}

The solutions to $x$ ‍ in the interval from $- π$ ‍ to $π$ ‍ are $x \approx 1.49$ ‍ and $x \approx 1.65$ ‍.

What was Jamie's mistake?

(Choice A)
Jamie evaluated $\cos^{- 1} (- 0.08)$ ‍ in degrees instead of in radians.
(Choice B)
Jamie used an identity that only applies to the sine function, not to the cosine function.
(Choice C)
Jamie forgot to use negative $n$ ‍-values which would reveal more solutions in the interval.
(Choice D)
Jamie gave solutions that were outside of the interval.

Stuck?