SOLUTION: Find all values of x in the interval [0, 2π] that satisfy the given equation. (sin x)^2 - 12 cos x - 12 = 0

Algebra ->  Trigonometry-basics -> SOLUTION: Find all values of x in the interval [0, 2π] that satisfy the given equation. (sin x)^2 - 12 cos x - 12 = 0      Log On


   

Question 808521: Find all values of x in the interval [0, 2π] that satisfy the given equation.

(sin x)^2 - 12 cos x - 12 = 0
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
(1) sin%5E2%28x%29+-+12cosx+-12+=+0
Use the trig identity
(2) sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+1 to get
(3) sin%5E2%28x%29+=+1+-+cos%5E2%28x%29
Put (3) into (1) to obtain
(4) 1+-+cos%5E2%28x%29+-+12cosx+-12+=+0 or
(5) -+cos%5E2%28x%29+-+12cosx+-+11+=+0 or
(6) cos%5E2%28x%29+%2B+12cosx+%2B+11+=+0
which factors into
(7)%28cos%28x%29+%2B+1%29%2A%28cosx+%2B+11%29+=+0
Now set each factor equal to zero and solve for x, and get
(8)cos%28x%29+%2B+11+=+0 and
(9)cosx+%2B+1+=+0 or
(10)cosx+=+-11 and
(11)cosx+=+-1
Since absolute value of cos(x) is never greater than one, (10) has no solutions, whereas (12) is
(13) x = arccos(-1) or
(14) x = 180 degrees
Let's check this with (1).
Is (sin(180)^2 -12cos(180) -12 = 0)?
Is (0^2 - 12*(-1) - 12 = 0)?
Is (0 + 12 - 12 = 0)?
Is (0 = 0)? Yes
Answer: the only x in the interval [0,360] is x = 180 degrees or pi radians.