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Marat540 [252]
4 years ago
7

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

Mathematics
2 answers:
svetoff [14.1K]4 years ago
8 0

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.

photoshop1234 [79]4 years ago
5 0

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.

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