Solve the following equation for angles between 0 and 360 degrees.?

Question

(3sin^2 x) + (cos x) = 1

10 points to whoever provides the most concise step-by-step answer.

Answer 1

sin^2 x=1-cos^2 x
(3sin^2 x) + (cos x) = 1
3(1-cos^2 x)+(cos x)=1
rearrange, we get
3 cos^2 x-cos x-2=0
solve the quadratic equation,
(3 cos x+2)(cos x-1)=0
cos x =-2/3 or 1
x=0,131.8,228.2,360(all in degree)

Answer 2

Since cos^2 x+ sin^2 x = 1, I would convert the equation to an equation in cosine form only:
3(1- cos^2(x) ) + cos x = 1.
Now you can form a friendly quadratic in terms of Cos x [3cos^x- cosx-2=0], and solve it,(if U=cosx,
3U^2-U-2=0) rejecting any roots that are imaginery or greater than 1. This looks like it factors nicely to (3U+2)(U-1)=0. and you will have 4 answers.

Answer 3

remember this identity from trig: sin²(x) = 1 - cos²(x)

3sin²(x) + cos(x) =1
3(1 - cos²(x)) + cos(x) = 1
3 - 3cos²(x) + cos(x) = 1
3cos²(x) - cos(x) + 2 = 0

let z = cos(x)

3z² - z - 2 = 0

(3z + 2) (x - 1) = 0

this means z = 1 and z = -2/3 which in turn means
cos(x) = 1 and cos(x) = -2/3

arccos(1) = 0° and 360°
arccos(-2/3) = 131.81° and 228.19°

Answer 4

x=0 or x= cos^-1(-0.66)

I think u can find the answer,
4 next time do ur homework at home!!

Answer 5

Buy a TI-89 and use the eqn. solver.

Or write and use a root finding algorithm like newton's method.