# Solving trigonometric equations

• May 11th 2011, 04:19 PM
KevRod43
Solving trigonometric equations
So hey guys this is college precal, I need help with two questions. can you help me figure them out? :]

1. Find all solutions of Sin(4x) = \frac{1}{\sqrt{2} } in degrees.

2. Find all solutions of Cos(2x-\frac{\pi }{ 6}) = \frac{\sqrt{3} }{2 }
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Any help is appreciated thanks!
• May 11th 2011, 05:20 PM
Soroban
Hello, KevRod43!

Quote:

$\displaystyle \text{1. Find all solutions in degrees: }\;\sin(4x) \:=\: \frac{1}{\sqrt{2}}$

We have: .$\displaystyle 4x \:=\:\begin{Bmatrix}45^o + 360^on \\ 135^o + 360^on \end{Bmatrix}$

Therefore: .$\displaystyle x \;=\;\begin{Bmatrix}\frac{45}{4}^o + 90^on \\ \\[-3mm] \frac{135}{4}^o + 90^on\end{Bmatrix}\;\text{ for any integer }n.$

Quote:

$\displaystyle \text{2. Find all solutions of: }\;\cos(2x-\tfrac{\pi}{6}) \:=\: \frac{\sqrt{3}}{2}$

We have: .$\displaystyle 2x - \tfrac{\pi}{6} \;=\;\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-3mm]\text{-}\frac{\pi}{6}+ 2\pi n\end{Bmatrix}$

. . . . . . . . . . . $\displaystyle 2x \;=\;\begin{Bmatrix}\frac{\pi}{6} + \frac{\pi}{6} + 2\pi n \\ \\[-3mm] \frac{\pi}{6} - \frac{\pi}{6} + 2\pi n \end{Bmatrix} \;=\;\begin{Bmatrix}\frac{\pi}{3} + 2\pi n \\ \\[-3mm] 2\pi n\end{Bmatrix}$

. . . . . . . . . . . .$\displaystyle x \;=\;\begin{Bmatrix}\frac{\pi}{6} + \pi n \\ \pi n\end{Bmatrix}\;\text{ for any integer }n$