Is the following statement true or false:

If asked to solve for x in the following equation:

`(1+sinx)/(cosx) + (cosx)/(1+sinx) = 4`

For `0<=x<=2Pi`

`x = Pi/2` ?

### 2 Answers | Add Yours

`(1+sin(x))/cos(x)+cos(x)/(1+sin(x))=4`

`((1+sin(x))^2+cos^2(x))/(cos(x)(1+sin(x)))=4`

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

`2+2sin(x)=4cos(x)(1+sin(x))`

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

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

`either`

`1+sin(x)=0`

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

`x=(3pi)/2`

`r`

`4cos(x)=2`

`cos(x)=1/2`

`cos(x)=cos(pi/3)`

`x=pi/3`

``So your answer is not correct or **false **

Solve `(1+sinx)/(cosx)+(cosx)/(1+sinx)=4` :

`(1+sinx)/(cosx)+(cosx)/(1+sinx)=4`

`((1+sinx)(1+sinx)+cos^2x)/(cosx(1+sinx))=4`

`sin^2x+2sinx+1+cos^2x=4cosx(1+sinx)`

`2sinx+2=4cosx(1+sinx)`

`2(sinx+1)=4cosx(1+sinx)`

If you divide both sides by 1+sinx you get:

`4cosx=2 ==> cosx=1/2 ==> x=pi/3,(5pi)/3` for `0<=x<=2pi`

However, these may not be the only solutions. If 1+sinx=0 then this equation holds. (Dividing by zero is not allowed, so this might represent a lost root.)

1+sinx=0 ==> sinx=-1 ==> `x=(3pi)/2`

But this solution does not work in the original equation as `cos((3pi)/2)=0` and you cannot divide by 0.

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The solutions are `x=pi/3,(5pi)/3`

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The graph:

Note that your solution is also impossible as `cos(pi/2)=0` and you would be dividing by zero in the first term.

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