I am stuck on the following problem since I can only get one of the solutions I want:
Solve $\cos(x-60)◦ = 0.6$, in the interval $0 \leq x \leq 360◦$, to one decimal place.
What I kind of understand so far:
$\cos(x^o) \mapsto cos(x^o-60^o)$ is a transformation in the x direction by $60^o$, so the new interval is $60^o ≤ x^o+60^o ≤ 420◦$?
I rearrange so that $x^o-60^o = \cos^{-1}(0.6)$, and solve so that $x = 113.1◦$.
Our teacher told us that when we get the $x$ value for $\cos$, we always do $360$ subtract that value to get the next solution, but that gave me the wrong answer $(246.9◦)$.
Answers:
$x = 6.9◦, 113.1◦$
Could someone explain how the answer is $6.9◦$? Thanks
Best Answer
Just taking the inverse cosine of the result is the "naive" way of solving the equation because it reveals only one solution out of an infinity of solutions. The proper way to solve this is to remember that $$ \cos{x} = \cos{y} \qquad \Leftrightarrow \qquad x=\pm y + n2\pi, \qquad n\in \mathbb{Z} $$ Therefore, you have to include the plus/minus after taking the inverse cosine. So as an "exam answer" I would write the following: $$ \cos{(x-60^{\circ})}= 0.6 \qquad \Rightarrow \qquad x-60^{\circ} \approx \pm 53.13^{\circ} + n2\pi , n\in \mathbb{Z} $$ and therefore $x\approx 53.13^{\circ}+60^{\circ} \approx 113.13^{\circ}$ or $x= -53.13^{\circ}+60^{\circ}\approx 6.87^{\circ}$. These are the two solutions in the defined range.