Solve, for $0° < x < 180°$, the equation $\sin(2x + 50°) = 0.6$

Question

I can't get the right answer. Manipulating the equation algebraically I get $x = -6.6$

From my understanding, if we graphed this on the x and y axis, we would shift the curve -50 and compress it so that the period is 90° instead of 180°.

I'm getting completely wrong answers so there are probably several flaws in my logic here.. can someone point them out?

Thanks in advance

Answer 1

Well, because you require a value for the angle $x$ between $0^\circ$ and $180^\circ$, just add $180^\circ$ to your answer to get $173.4^\circ$.

Why is this valid? That's because the sine function is repeatedly curved, it will have the same value at regular intervals.

In response to your comment, we actually observe that the solutions to the equation $\sin x = a$ (where $- 1 \leq x \leq 1$) are: $$x = \arcsin a + 2k\pi$$ $$x = \pi - \arcsin a + 2k\pi$$ where $k \in \mathbf{Z} $. For your information, all the above angles are expressed in radians. For more explanation, you can also see here.

Hence, we have as another solution: $$ 2x + 50 = 180^\circ - 36.9^\circ$$ giving the required second answer as: $x = 46.6^\circ$.