# What is the answer for question 2) ? http://postimg.org/image/con6jawnr/

embizze | Certified Educator

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Solve `2cos^2x-7cosx+3=0` on the interval `0<=x<=2pi`

Factor the left side. (This is a quadratic in cosx)

`(2cosx-1)(cosx-3)=0`

By the zero product property, at least one of the factors must be zero.

(1) `2cosx-1=0 ==gt cosx=1/2`

`cosx=1/2 ==gt x=pi/3,(5pi)/3`

(2) cosx-3=0 ==> cosx=3 which is impossible. (The range for cosx is -1 to 1.)

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The solution is `x=pi/3,x=(5pi)/3`

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The graph:

embizze | Certified Educator

Solve the equation `-sin^2x=cos^2x`on the interval `0<=x<=2pi`

`sin^2x+cos^2x=0`

But by the Pythagorean identity we know that `sin^2x+cos^2x=1`

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There is no solution

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The graph of `y=-sin^2x` in black, `y=cos^2x` in red: