<s>[INST] solve $|\sin(a)|=\cos(3a)$ $|\sin(a)|=\cos(3a)$ is an alternative version of an equation $\sqrt{1-x^2}=4x^3-3x$, where I made a substitution $x=\cos(a)$ for $x \in [-1, 1]$. Unfortunately, I have no idea how to solve trigonometric equations.
 [/INST] Consider
Case 1:  $0 \le a \le \frac \pi 6$ then $0 \le 3a \le \frac \pi 2$.  Than $\sin a \ge 0$ and $\cos 3a \ge 0$.
Then $a$ and $3a$ are both in the first quandrant.  And the identity that will apply is $\sin a = \cos (\frac \pi 2 - a)$.
So we must have $\frac\pi 2 - a= 3a$ or $a =\frac {\pi} 8$.
Case 2:  $\frac \pi 6 < a < \frac \pi 2$.  Then $\frac \pi 2 < 3a < \frac {3\pi}2$.  But then $\cos 3a < 0$ and that's not the case.
Case 3: $\frac \pi 2 \le a \le \frac {2\pi}3$.  And $\frac {3\pi}2 \le 3a \le 2\pi$.  Then $\sin a \ge 0$ and $\cos 3a \ge 0$.
Now $a$ is in the second quadrant and $a$ is in the fourth and the indentity that is relevent is.
$\sin (\frac \pi 2 + \alpha) = \cos (2\pi - \alpha)$.
So $a = \frac \pi 2 + \alpha$ and $3a = 2\pi - \alpha$ so $a+ 3a = 2\pi + \frac \pi 2$ so $4a = \frac {5\pi }2$ and $a = \frac {5\pi} 8$.
Case 4: $\frac {2\pi}3 < a \le  \frac {5\pi}6$ and $2\pi < 3a \le 2\pi + \frac {\pi}2$.
Then $a$ is in second quadrant but $3a$ is in the first but over $2\pi$.
The identity that applies here is $\cos (2\pi + \alpha) = \sin (\frac \pi 2 + \alpha)$.
So $a = \frac \pi 2 + \alpha$ and $3a = 2\pi + \alpha$ ans so $3a - a = \frac {3\pi }2$ and $a= \frac {3\pi}4$.
Case 5: $\frac {5\pi}6 < a < \frac {7\pi}6$ and $\frac {5\pi}2 < 3a < \frac {7\pi}2$.  But then $\cos 3a < 0$ which isn't possible.
Case 6: $\frac {7\pi}6 \le a \le \frac {4\pi}3$ so $\frac {7\pi}2 \le 3a \le 4\pi$ then $\sin a \le 0$ but $\cos 3a = -\sin a \ge 0$.
The identity here is $-\sin (\pi + \alpha) = \sin (\pi - \alpha)=\cos(\frac {7\pi}2 + \alpha)$.
So $a = \pi + \alpha$ and $3a = \frac {7\pi}2 + \alpha $ so $2a = \frac {5\pi}2$ and $a =\frac {5\pi}4$.
Case 7: $\frac {4\pi}3 \le a \le \frac {3\pi}2$ so $4\pi \le 3a \le \frac {9\pi}2$.
Then identity to use here is $-\sin (\frac {3\pi} 2 - \alpha)= \cos (4\pi + \alpha)$
So $a = \frac {3\pi}2  -\alpha$ and $3a =4\pi +\alpha$ so $4a = 4\pi + \frac {3\pi}2$ so $a =\pi +\frac {3\pi}8 = \frac {11\pi}8$.
Case 8: $\frac {3\pi}2 < a < \frac {11\pi } 6$ so $\frac {9\pi}2< 3a < \frac {11\pi}2$ so $\cos 3a < 0$.
And final final case
9:  $\frac {11\pi}6 \le a < 2\pi$ then $\frac {11\pi}2 \le 3a < 6\pi$.
The identity we need is $-\sin (\frac {3\pi}2 + \alpha) = \cos(6\pi - \alpha)$.
So $a + 3a = 6\pi + \frac {3\pi }2$ and so $a = \frac {15\pi}8$.
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