Question

Find all solutions to the equation (sinx-cosx)^2=3

Answer 1

Answer:

no solutions

Step-by-step explanation:

Simplify the equation $(\sin x-\cos x)^2=3:$

$\sin^2 x-2\sin x\cos x+\cos ^2x=3,\\ \\ (\sin^2x+\cos^2x)-2\sin x\cos x=3.$

Since

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

and

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

we have

$1-\sin 2x=3,\\ \\\sin 2x=-2.$

Since $\sin 2x$ cannot be less than $-1,$ this equation has no solutions.

Answer 2

Answer:

This equation has no solution.

Step-by-step explanation:

We have given the equation:

(sinx-cosx)²= 3

We have to solve it.

(sinx-cosx)²= 3

sin²x+cos²x-2sinxcosx = 3

As we know that :

sin²x+cos²x = 1

so, (1)-2sinxcosx = 3

1-2sinxcosx = 3

1-sin2x = 3

sin2x=1-3

sin2x = -2

2x = sin⁻¹(-2)

The value of sinx cannot  be less than -1 so, this equation has no solution.