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

Question

1699 views around the world

You can reuse this answer
Creative Commons License

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 $[0, 2pi], x = 0. pi/2 and 2pi$

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