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What is the answer for question 2) http://postimg.org/image/jjjnw5hzr/ Reminder: This...

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lkballer24 | Student, Grade 11 | Valedictorian

Posted August 18, 2013 at 11:12 PM via web

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What is the answer for question 2)

http://postimg.org/image/jjjnw5hzr/

Reminder: This is one question.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 19, 2013 at 1:10 AM (Answer #1)

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Solve for `0<=x<=2pi` :

(1) `cos^2x-3/4=0`

`cos^2x=3/4`

`cosx=+-sqrt(3)/2`

`x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6`

(2) `sin^2x=cos^2x`

`tan^2x=1`

`tanx=+-1`

`x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4`

(3) `2cos^2x-7cosx+3=0`

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

`2cosx-1=0 ==> cosx=1/2 ==> x=pi/3,(5pi)/3`

`cosx-3=0 ==> cosx=3` which is impossible.

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