# Solving a trig equation

• Nov 24th 2009, 10:10 AM
Solving a trig equation
Find, to the nearest degree, all values of theta in the interval 0° ≤ theta < 360°
that satisfy the equation 3 cos 2theta + sin theta – 1 = 0.
• Nov 24th 2009, 11:43 AM
apcalculus
At:

Wolfram|Alpha

enter the command:

plot 3 cos(2x) + sin(2x) - 1 for x between 0 and 360 degrees

it looks like the equation has four zeros in your interval. Use your graphing calculator to approximate them.

Good luck!
• Nov 24th 2009, 01:39 PM
Soroban

Quote:

Find, to the nearest degree, all values of $\displaystyle \theta$ in the interval $\displaystyle [0^o,\:360^o)$

that satisfy the equation: .$\displaystyle 3\cos2\theta + \sin\theta – 1 \:=\: 0$

Use the identity: .$\displaystyle \cos2\theta \:=\:1-2\sin^2\!\theta$

The equation becomes: .$\displaystyle 3(1-2\sin^2\!\theta) + \sin\theta - 1 \:=\:0 \quad\Rightarrow\quad 6\sin^2\!\theta - \sin\theta - 2 \:=\:0$

Factor: .$\displaystyle (2\sin\theta + 1)(3\sin\theta - 2) \:=\:0$

And we have two equations to solve:

. . $\displaystyle 2\sin\theta + 1 \:=\:0 \quad\Rightarrow\quad \sin\theta\:=\:\text{-}\tfrac{1}{2} \quad\Rightarrow\quad \theta \:=\:210^o,\:330^o$

. . $\displaystyle 3\sin\theta -2 \:=\:0 \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{2}{3} \quad\Rightarrow\quad \theta \:\approx\:42^o,\:138^o$