SOLUTION: Find all values of x in the interval [0, 2π] that satisfy the given equation.
(sin x)^2 - 12 cos x - 12 = 0
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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) (Show Source): You can put this solution on YOUR website!
Given:
(1)
Use the trig identity
(2) to get
(3)
Put (3) into (1) to obtain
(4) or
(5) or
(6)
which factors into
(7)
Now set each factor equal to zero and solve for x, and get
(8) and
(9) or
(10) and
(11)
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 radians.
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