Can someone solve this trig equation?

Question

Can someone solve the following trig equation:-
sin^2(x) = 4cos(x)

Thanks

Answer 1

sin^2 x = 4 cos x
1 - cos^2 x = 4 cos x
0 = cos^2 x + 4 cos x - 1

Let y = cos x

0 = y^2 + 4y - 1

Using quadratic formula:
y = [-4 +- sqrt (4^2 - 4*1*-1)] / 2(1)
y = [-4 +- sqrt 20] / 2
y = [-4 +- 2sqrt 5] / 2
y = -2 +- sqrt 5
y = -4.236 and 0.236

so...
cos x = -4.236 and 0.236

we can elminate the -4.236 because cos x is always between -1 and 1.

So...
cos x = 0.236
x = 76.345 degrees (approx)

Answer 2

Put 1 - cos^2 in place of sin^2, then let m = cos x. This will leave you with the quadratic equation
m^2 + 4m - 1 = 0. Solve for m then put cos(x) back in place and solve for x.

Answer 3

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Answer 4

1-cos²x=4cos(x)
cos²x+4cos(x)-1=0
cos(x)=(-4±√(16+4))/2
cos(x)=-2±√5
cos(x)=-4.2360,.2360
obviously only one possible
cos(x)=.2360
x=76.3

Answer 5

+atan(2/(-2+5^(1/2))^(1/2)),
-atan(2/(-2+5^(1/2))^(1/2))
+atan(2*(-2-5^(1/2))^(1/2),-2-5^(1/2))
-atan(-2*(-2-5^(1/2))^(1/2),-2-5^(1/2))