#### Question

# The solutions for the equation 2cos^2 x - 3cos^2 x + 1= 0 for -pi<x<pi ?

With working and explanation please? :(

#### Answers

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

-->

You have like terms:

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

-->

-cos2(x) + 1 = 0

--> solve for cos2(x)

cos2(x) = 1

-->

cos(x) = ±√1 = ±1

Where does cos(x) = 1 and -1?

cos(x) = 1 --> (1, 0) --> the angle is 0 --> x = 0

cos(x) = -1 00> (-1, 0) --> angle = π or 180°

...technically that point isn't in the range you gave, so the only solution is x = 0...

but, you probably meant one of the following:

-π ≤ x < π --> x = -π

-π < x ≤ π --> x = +π

So the answer is either:

x = 0, π

or

x = 0, -π

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