SOLUTION: Find all degree solutions in the interval 0° &#8804; &#952; < 360°. If rounding is necessary, round to the nearest tenth of a degree. 6sin^2(&#952;) &#8722; 7cos2(&#952;) = 0

Algebra ->  Trigonometry-basics -> SOLUTION: Find all degree solutions in the interval 0° &#8804; &#952; < 360°. If rounding is necessary, round to the nearest tenth of a degree. 6sin^2(&#952;) &#8722; 7cos2(&#952;) = 0       Log On


   

Question 1004017: Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree.
6sin^2(θ) − 7cos2(θ) = 0
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
From 6sin^2(θ) − 7cos^2(θ) = 0
6sin^2(θ) − 7(1 - sin^2(θ) = 0
13sin^2(θ) − 7 = 0
13sin^2(θ) = 7
sin^2(θ) = 7/13
sin(θ) = +/- sqrt(7/13) = +/- .738
so that
θ = 47.2, 132.8, 227.2, 312.8 degrees