Question

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

Answer 1

x = πn/2

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.

Answer 2

`x=(\pi k)/(2),\ k\in Z.`

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=(\pi k)/(2),\ k\in Z.`