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If asked to solve for x for the following: `sin(x)tan(x) +tan(x) - 2sin(x) + cos(x) =...

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foxwit | Student, Undergraduate | Salutatorian

Posted July 27, 2013 at 5:44 PM via web

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If asked to solve for x for the following:

`sin(x)tan(x) +tan(x) - 2sin(x) + cos(x) = 0`

are the following solutions correct:

`x = (Pi)/(2) , and (3Pi)/(2)` 

1 Answer | Add Yours

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aruv | High School Teacher | Valedictorian

Posted July 27, 2013 at 6:13 PM (Answer #1)

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`sin(x)tan(x)+tan(x)-2sin(x)+cos(x)=0`

`sin(x)(sin(x))/cos(x)+(sin(x))/(cos(x))-2sin(x)+cos(x)=0`

`sin^2(x)+sin(x)-2sin(x)cos(x)+cos^2(x)=0`

`sin^2(x)+cos^2(x)+sin(x)-2sin(x)cos(x)=0`

`1+sin(x)-2sin(x)cos(x)=0`

`(1+sin(x))^2=(2sin(x)sqrt(1-sin^2(x)))^2`

`1+sin^2(x)+2sin(x)=2sin^2(x)(1-sin^2(x))`

`1+sin^2(x)+2sin(x)=2sin^2(x)-2sin^4(x)`

`2sin^4(x)-sin^2(x)+2sin(x)+1=0`

`2sin^4(x)+2sin^3(x)-2sin^3(x)-2sin^2(x)+sin^2(x)+sin(x)+sin(x)+1=0`

`(sin(x)+1)(2sin^3(x)-2sin^2(x)+sin(x)+1)=0`

`either`

`sin(x)+1=0`

`sin(x)=-1=sin((3pi)/2)`

`x=(3pi)/2`

`or`

`2sin^3(x)-2sin^2(x)+sin(x)+1=0`    (i)

So your one aswer is correct x=(3pi)/2

and x=pi/2 is not correct because it will not satisfy (i).

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