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

1 Answer
Sep 28, 2016

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

Explanation:

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