# Trig question: with intervals

• Jan 13th 2009, 07:51 PM
VkL
Trig question: with intervals
Solve the equation for a in [0, 2pie]

A.) cosē(a) + 3cos(a) = -cos(a)

B.)sin(2a) = cos(a)
• Jan 13th 2009, 08:08 PM
ThePerfectHacker
Quote:

Originally Posted by VkL
Solve the equation for a in [0, 2pie]

A.) cosē(a) + 3cos(a) = -cos(a)

Add $\displaystyle \cos a$ to both sides to get $\displaystyle \cos^2 a + 4\cos a = 0$ and now $\displaystyle \cos a ( \cos a + 4) = 0$. Therefore, $\displaystyle \cos a = 0$ or $\displaystyle \cos a + 4 = 0$. The second condition, $\displaystyle \cos a + 4 = 0 \implies \cos a = -4$ is impossible because $\displaystyle |\cos a| \leq 1$. The first condition is $\displaystyle \cos a = 0$. That happens when $\displaystyle a = \tfrac{\pi}{2},\tfrac{3\pi}{2}$

Quote:

B.)sin(2a) = cos(a)
Using the double angle identity we get $\displaystyle 2\sin a \cos a = \cos a \implies 2\sin a \cos a - \cos a = 0$. Now factor, $\displaystyle \cos a ( 2\sin a - 1) = 0$. Thus, we get two possibilities, $\displaystyle \cos a = 0$ or $\displaystyle 2\sin a - 1 = 0$. The first condition, $\displaystyle \cos a = 0$, gives $\displaystyle a = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$. The second condition, $\displaystyle 2\sin a - 1 = 0$ is equivalent to saying $\displaystyle \sin a = \tfrac{1}{2}$, thus, $\displaystyle a = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$.