# Solve the equation for all values of x between 0 and 180 degrees. 2sin^2 x + 3cos x = 0 .

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### 2 Answers

We have to solve 2*(sin x)^2 + 3*cos x = 0 for values of x in the interval {0, 180}

2*(sin x)^2 + 3*cos x = 0

=> 2*(1 - (cos x)^2) + 3*cos x = 0

=> 2 - 2*(cos x)^2 + 3*cos x = 0

=> 2*(cos x)^2 - 3*cos x - 2 = 0

=> 2*(cos x)^2 - 4*cos x + cos x - 2 = 0

=> 2*(cos x)[ cos x - 2] + 1[cos x - 2] = 0

=> [2*(cos x) + 1][ cos x - 2] = 0

[2*(cos x) + 1] = 0

=> 2*cos x = -1

=> cos x = -1/2

=> x = arc cos(-1/2)

=> x = 120 degrees

cos x - 2 = 0

=> cos x = 2 which is not possible

**The required solution of the equation is x = 120 degrees**

We'll use the Pythagorean identity:

(sin x)^2 = 1 - (cos x)^2

We'll re-write the equation:

2 - 2(cos x)^2 + 3cos x = 0

2(cos x)^2 - 3cos x - 2 = 0

We'll replace cos x by t:

2t^2 - 3t - 2 = 0

We'll apply quadratic formula:

t1 = [3+sqrt(9+16)]/4

t1 = (3+5)/4

t1 = 2

t2 = -1/2

But cos x = t1 => cos x = 2 impossible

cos x = t2 => cos x = -1/2

Since cosine function is negative in the second quadrant, x = pi - pi/3 = 2pi/3 = 120 degrees.