SOLUTION: Solve for 0 < theta < 2pi, giving your answers in terms of pi. 1. Tan theta = - "square root 3 over 3" 2. Cos^2 theta = 1 over 4

Algebra ->  Trigonometry-basics -> SOLUTION: Solve for 0 < theta < 2pi, giving your answers in terms of pi. 1. Tan theta = - "square root 3 over 3" 2. Cos^2 theta = 1 over 4      Log On


   



Question 1043597: Solve for 0 < theta < 2pi, giving your answers in terms of pi.
1. Tan theta = - "square root 3 over 3"
2. Cos^2 theta = 1 over 4

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Triangle that is 1-sqrt3-2 has a tangent that is 1/sqrt(3), and that is sqrt(3)/3
That triangle is a 30-60-90 triangle, so the tangent that gives that value is 30 degrees or pi/6 radians.
It also occurs when the sine and cosine are negative (tangent=sin/cos), and that would be at 210 degrees or 7pi/6 radians.
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cos^2 theta=1/4
cos theta= +/- 1/2
That occurs at pi/6, pi/3, 2pi/3, and 5pi/3. Notice that -1/2 is a solution, so there are twice as many instances.
30-120-240-300 degrees.