How do you solve $3(1-sinx)=2cos^2x$ in the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-11-09T22:36:40

$3 - 3sinx = 2cos^2x$

Apply the identity $sin^2theta + cos^2theta = 1- > cos^2theta = 1 - sin^2theta$ .

$3 - 3sinx = 2(1 - sin^2x)$

$3 - 3sinx = 2 - 2sin^2x$

$2sin^2x - 3sinx + 1 = 0$

$2sin^2x - 2sinx - sinx + 1 = 0$

$2sinx(sinx - 1) - (sinx - 1) = 0$

$(2sinx - 1)(sinx - 1) = 0$

$sinx = 1/2 and sinx = 1$

$x = pi/6, (5pi)/6, pi/2$

Hopefully this helps!

How do you solve 3(1-sinx)=2cos^2x3(1-sinx)=2cos^2x in the interval 0<=x<=2pi0<=x<=2pi?