How do you solve $3 (1 - sin x) = 2 {cos}^{2} x$ $3(1-sinx)=2cos^2x$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

Question

5626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T22:36:40

$3 - 3 sin x = 2 {cos}^{2} x$ $3 - 3sinx = 2cos^2x$

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

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

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

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

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

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

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

$sin x = \frac{1}{2} and sin x = 1$ $sinx = 1/2 and sinx = 1$

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{π}{2}$ $x = pi/6, (5pi)/6, pi/2$

Hopefully this helps!

How do you solve 3(1-sinx)=2cos^2x3(1−sinx)=2cos2x3(1-sinx)=2cos^2x in the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?