# What is the answer for question 1) ? http://postimg.org/image/con6jawnr/

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### 2 Answers

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Solve `-cos^2x=-cosx` on the interval `0<=x<=360^@` :

Note: do not divide both sides by cosx; this will result in a lost root. Add `cos^2x` to both sides to get:

`cos^2x-cosx=0`

`cosx(cosx-1)=0`

Either cosx=0 ==> `x=90^@,270^@`

or `cosx-1=0 ==> cosx=1 ==> x=0,360^@`

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The solutions are `x=0,90^@,270^@,360^@`

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

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Solve `6sin^2x-11sinx+4=0` on the interval `0<=x<=360^@` . (This is quadratic in sinx.)

Factor the left side:

`(3sinx-4)(2sinx-1)=0`

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

(1) `2sinx-1=0 ==gt sinx=1/2 ==gt x=30^@,x=150^@`

(2) `3sinx-4=0 ==gt sinx=4/3` which is impossible as the range for the sine is -1 to 1.

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` x=30^@,x=150^@` are the solutions

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