Solve the equation for a in [0, 2pie]

A.) cosē(a) + 3cos(a) = -cos(a)

B.)sin(2a) = cos(a)

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- Jan 13th 2009, 07:51 PMVkLTrig 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 PMThePerfectHacker
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 $\displaystyle a = \tfrac{\pi}{2},\tfrac{3\pi}{2}$

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B.)sin(2a) = cos(a)