# Solving $\sin3v\lt \frac{1}{2},\{0^o\lt v \lt 90^{o}\}$ wrongly, how do I solve it correctly?

For which angles $0^o\lt v \lt 90^{o}$ is it true that $\sin3v\lt \frac{1}{2}$?

Solve $\sin3v\lt \frac{1}{2},\{0^o\lt v \lt 90^{o}\}$.

$\sin3v\lt \frac{1}{2}$ gives that

$$\left\{\begin{matrix} & v_1\lt 10 \\ & v_2\lt 50 \end{matrix}\right.$$ and here it got wrong.

Clearly $\sin3v$ decreases for $v\gt 50$, but I got that $v_2\lt50$ in the given interval. I did not divide or multiply with a negative variable, so what did I do wrong?

EDIT: What I want is an algebraic clarification of what I did wrong when solving the inequality.

• You should have $v_2 > 50^\circ$. – N. F. Taussig May 7 '18 at 10:20
• @N.F.Taussig, correct, but why? – Andreas May 7 '18 at 10:21

Let $u = 3v$. Then $0^\circ < v < 90^\circ \implies 0^\circ < u < 270^\circ$. Hence, your question is equivalent to asking for which angles satisfying $0^\circ < u < 270^\circ$ is $\sin u < \frac{1}{2}$?
We know that $\sin u$ increases from $0$ to $1$ as $u$ increases from $0$ to $\pi/2$ ($0^\circ$ to $90^\circ$) and is equal to $1/2$ when $u = \pi/6$ ($30^\circ$). It then decreases from $1$ to $-1$ as $u$ increases from $\pi/2$ to $3\pi/2$ ($90^\circ$ to $270^\circ$) and is equal to $1/2$ when $u = 5\pi/6$ ($150^\circ$).
Hence, the inequality $\sin u < 1/2$ is satisfied when $0 < u < \pi/6$ ($0^\circ < u < 30^\circ$) or $5\pi/6 < u < 3\pi/2$ ($150^\circ < u < 270^\circ$). Since $u = 3v$, the inequality $\sin 3v < 1/2$ is satisfied when $0^\circ < v < 10^\circ$ or $50^\circ < v < 90^\circ$.
Addendum: We know that $\sin(3v) = \frac{1}{2}$ when $v = 10^\circ$ or $v = 50^\circ$. Since $\sin(3v)$ is increasing as $v$ increases from $0^\circ$ to $30^\circ$, the sign of $\sin(3v) - 1/2$ changes from negative to positive at $10^\circ$. Since $\sin(3v)$ is decreasing as $v$ increases from $50^\circ$ to $90^\circ$, the sign of $\sin(3v) - 1/2$ changes from positive to negative at $50^\circ$. The inequality $\sin(3v) < 1/2$ is satisfied when $\sin(3v) - 1/2 < 0$.