Question

Can you help me with this problem: `sinx+cosx=sinxcosx`? Please?

Answer 1

`x=pi+1/2arcsin(2-2sqrt2)`

Explanation:

`sinx+cosx=sinxcosx`

`(sinx+cosx)^2=sin^2xcos^2x`

`sin^2x+2sinxcosx+cos^2x=sin^2xcos^2x`

`1+2sinxcosx=sin^2xcos^2x`

Let `y=sinxcosx`

`1+2y=y^2`

`y^2-2y-1=0`

`y=1+-sqrt2`

`sinxcosx=1+-sqrt2`

`sin2x = 2sinxcosx therefore 1/2sin2x=sinxcosx`

`1/2sin2x=1+-sqrt2`

`sin2x=2+-2sqrt2`

`2+2sqrt2>1 therefore` it has no solutions

`sin2x=2-2sqrt2`

`2x=arcsin(2-sqrt2)`

`x=1/2arcsin(2-sqrt2)`

Checking this value, we see that the two sides of the equation don't have the same value. This is because `sinx <0` but `abs(cosx)>abs(sinx)`, so one side of the eqn is positive whilst the other is negative.

So we add `pi` to the value to make `sin` positive and `cos` negative but we also keep `abs(cosx)>abs(sinx)`, so now both sides are negative and, more importantly, have the same value.

So `x=pi+1/2arcsin(2-2sqrt2)`

Answer 2

`x = pi+1/2 arcsin(2(1-sqrt(2)))`

Explanation:

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

so

`(sinx cosx)^2-2sinxcosx-1=0`

so

`sinx cosx = (2 pm sqrt(4+4))/2`

so

`2sinxcosx=2(1-sqrt(2))`

but

`sin(2x)=2sinx cosx = 2(1-sqrt(2))`

then

`x = pi+1/2 arcsin(2(1-sqrt(2)))`