Solve each equation for 0 ≤ theta < 2pi: 2 cos theta =- square root 3 sin theta + cos theta?

Question

solve each equation for 0 ≤ theta < 2pi: 2 cos theta =- square root 3 sin theta + cos theta?

5822 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-13T23:12:22

#{5pi}/6# and #{11pi}/6#

Trig as taught leans heavily on The Two Tired Triangles of Trig , 30/60/90 and 45/45/90. #sqrt{3}# is a hint we'll soon see the former.

Solve #2cos theta = - \sqrt{3} sin theta + cos theta#

There are two ways to go; let's solve for #tan theta# this time.

#cos theta = -\sqrt{3} sin theta #

#- 1/sqrt{3} = sin theta/cos theta = tan theta #

That's 30/60/90 in the second or fourth quadrant. The #sqrt{3}# the denominator means the long side is the cosine, so this is #150^circ.#

# tan theta = tan 150^circ #

#theta = 150^circ + 180^circ k quad # for integer #k#

Oops, I was supposed to be using radians.

#theta = {5pi}/6+ k pi quad # for integer #k#

In the range that's #{5pi}/6# and #{11pi}/6#.

Check:

#theta = {5pi}/6#

#2cos theta = 2 (-\sqrt{3}/2) = -sqrt{3} #

# - \sqrt{3} sin theta + cos theta = -\sqrt{3} (1/2) + - \sqrt{3}/2 = -sqrt{3} quad sqrt#

#{11pi}/6# negates all the terms so checks too.