sec (theta)=negative two over the square root of three
solve for theta
1
Expert's answer
2021-08-09T05:44:24-0400
secθ=−32
Recall:
secθ=cosθ1
Therefore,
cosθ1=−32cosθ=−23θ=cos−1(−23)θ=65π
The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from 2π to find the solution in the third quadrant.
θ=2π−65πθ=67π
The period of the cosθ function is 2π so values will repeat every 2π radians in both directions.
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