Question

Find all solutions to the equation (sinx-cosx)^2=1

Answer 1

Answer:

$x=\dfrac{\pi k}{2},\ k\in Z.$

Step-by-step explanation:

Simplify the equation $(\sin x-\cos x)^2=1:$

$\sin^2 x-2\sin x\cos x+\cos ^2x=1,\\ \\ (\sin^2x+\cos^2x)-2\sin x\cos x=1.$

Since

$\sin^2 x+\cos^2 x=1$

and

$2\sin x\cos x=\sin 2x,$

we have

$1-\sin 2x=1,\\ \\\sin 2x=0,\\ \\2x=\pi k,\ k\in Z,\\ \\x=\dfrac{\pi k}{2},\ k\in Z.$

Answer 2

Answer:

x = πn/2

Step-by-step explanation:

We have given the equation:

(sinx-cosx)²=1

We have to solve it.

(sinx-cosx)²=1

sin²x+cos²x-2sinxcosx =1

As we know that :

sin²x+cos²x = 1

so, (1)-2sinxcosx = 1

1-2sinxcosx = 1

1-sin2x =1

sin2x = 0

2x = sin⁻¹ (0)

Sinx is 0 at πk so we get,

2x = πn   (n belongs to Z)

x = πn/2 is the solution.