How do you solve #1= sin(x) + cos(x) + sin(x)cos(x)#?

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Answer 1 · 2016-09-28T13:37:00

The general solution is

#x=kpi+(-1)^n(pi/4)-pi/4, k = 0, +-1, +-2, +-3, ...#

In #[0, 2pi], x = 0, pi/2 and 2pi#

Use #sin x cos x = 1/2((sin x +cos x )^2-1)#.

Now, the given equation becomes

#(sin x + cos x)^2+2(sin x + cos x) - 3 = 0.#

This is a quadratic in

t = sin x + cos x

#= sqrt2(sin x/sqrt 2+ cos x /sqrt 2)#

#= sqrt2 sin (x +pi/4)#.

Solving, #t^2+2t-3=0#,

t = 1, -3.

-3 is inadmissible.

So, #sin(x+pi/4)=1/sqrt 2 = sin (pi/4).

The general solution is

#x=kpi+(-1)^n(pi/4)-pi/4, k = 0, +-1, +-2, +-3, ...#

In #[0, 2pi], x = 0. pi/2 and 2pi#