Math, asked by jag3, 1 year ago

7 sin^2 theta + 3 cos^2 theta =4, 0° <theta < 90°

Answers

Answered by vedha03
0
7sin^2Φ + 3cos^2Φ = 4
4sin^2Φ +3sin^2Φ +3cos^2Φ = 4
4sin^2Φ +3(sin^2Φ +cos^2Φ) = 4
4sin^2Φ +3 = 4
4sin^2Φ = 1
sin^2Φ = 1/4
sinΦ = +1/2 & -1/2
Φ = 30 and -30
here given that 0°<Φ<90°
so Φ = 30° = π/6 radians
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