Was I wrong to omit angles in the solution set for this multiple angle problem?

Question

I may have missed this in my precalculus course, but why was I wrong to omit angles that did not have a positive value for cosine? I didn't include $\frac{3\pi}{4},\frac{7\pi}{12},\frac{5\pi}{4}$ because I thought they would not equal positive $\frac{\sqrt{2}}{2}$. Here is a screenshot of my missed question. Thanks in advance!

Answer 1

You can be surer of not missing solutions by approaching the equation something like this. Designate the angle as $ \ \alpha \ = \ 3 \ x \ $ and solve the equation as

$$ \cos \alpha \ = \ \frac{\sqrt{2}}{2} \ \ \Rightarrow \ \ \alpha \ = \ \frac{\pi}{4} \ + \ 2 \ k \ \pi \ , \ \frac{7\pi}{4} \ + \ 2 \ k \ \pi \ \ . $$

We then have

$$ 3x \ = \ \frac{\pi}{4} \ + \ 2 \ k \ \pi \ , \ \frac{7\pi}{4} \ + \ 2 \ k \ \pi \ \ \Rightarrow \ \ x \ = \ \frac{\pi}{12} \ + \ \frac{2}{3} k \ \pi \ , \ \frac{7\pi}{12} \ + \ \frac{2}{3} k \ \pi $$

$$ \Rightarrow \ \ k \ = \ 0 \ : \ \ x \ = \ \frac{\pi}{12} \ , \ \frac{7 \pi}{12} \ \ ; $$

$$ \Rightarrow \ \ k \ = \ 1 \ : \ \ x \ = \ \frac{\pi}{12} \ + \ \frac{2 \cdot 1 \cdot \pi}{3} \ = \ \frac{9 \pi}{12} \ = \ \frac{3 \pi}{4} \ \ , $$ $$ \frac{7 \pi}{12} \ + \ \frac{2 \cdot 1 \cdot \pi}{3} \ = \ \frac{15 \pi}{12} \ = \ \frac{5 \pi}{4} \ \ ; $$

$$ \Rightarrow \ \ k \ = \ 2 \ : \ \ x \ = \ \frac{\pi}{12} \ + \ \frac{2 \cdot 2 \cdot \pi}{3} \ = \ \frac{17 \pi}{12} \ \ , \ \ \frac{7 \pi}{12} \ + \ \frac{2 \cdot 2 \cdot \pi}{3} \ = \ \frac{23 \pi}{12} \ \ . $$

You can stop at this point, since $ \ k = 3 \ $ will just add $ \ 2 \pi \ $ to your basic solutions; you have now covered the "principal circle" completely. A general "rule of thumb" is that a trigonometric equation with a multiple angle $ \ kx \ $ , such as $ \ \sin (kx) \ = \ C \ $ , has $ \ k \ $ times as many solutions in the principal circle as $ \ \sin x \ = \ C \ $ does (since the multiplier reduces the period of the trig function by a factor of $ \ k \ $ ) .

Answer 2

Well $\frac{3\pi}{4}$ is a solution in the sense they intended since it clearly is in $[0,2\pi)$ and $\cos(\frac{9\pi}{4}) = \frac{\sqrt{2}}{2}$.

Having said that I would just like to say I really dislike "problems" like this "homework question" whose sole purpose seems to be for people to make mistakes like this.