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

2 Answers
Monzur R.
Mar 17, 2017

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)

Cesareo R.
Apr 17, 2017

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)))

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