solve for x please: sin x - cos x = square root of 2?

Answer 1

sin x - cos x = √2
(sin x - cos x)² = 2
sin² x - 2(sin x)(cos x) + cos² x = 2
(sin² x + cos² x) - 2(sin x)(cos x) = 2
1 - 2(sin x)(cos x) = 2
-1 = 2(sin x)(cos x)
-1 = sin(2x)
2x = 3π/2
x = 3π/4 + 2πk radians

where k is an integer

Answer 2

Square both sides:

sin^2x + cos^2x - 2sinxcosx = 2
1 - 2sinxcosx = 2 (using identity sin^2 + cos^2 = 1)
2sinxcosx = -1
sin2x = -1/2 (double angle formula: sin2x=2sinxcosx)
2x = 3pi/2, 7pi/2, 11pi/2 etc...
x = 3pi/4, 7pi/4, 11pi/4, etc...

However, since we squared the original equation, we find only half of these are the correct solutions as some are the solutions of the equation sin x - cos x = - sqrt 2

Testing the solutions we see that the correct ones are 3pi/4, 11pi/4, etc and so the general solution is:

x = 2npi + 3pi/4, n = 0, 1, 2 .....

Answer 3

sin(x) - cos(x) = sqrt(2)

Square both sides, to get

sin^2(x) - 2sin(x)cos(x) + cos^2(x) = 2

Rearranging the terms a bit, we have

sin^2(x) + cos^2(x) - 2sin(x)cos(x) = 2

Note the famous identity, sine squared x plus cos squared x = 1.

1 - 2sin(x)cos(x) = 2
-2sin(x)cos(x) = 1
2sin(x)cos(x) = -1

Note the double angle identity sin2y = 2sinycosy.

sin2x = -1

Therefore, provided you have an interval restriction of [0, 2pi),

2x = {3pi/2, 7pi/2}
x = {3pi/4, 7pi/4}

Just being squaring values in an equation does crazy things sometimes, we should test the values back into the original equation.

sinx - cosx = sqrt(2)

Test x = 3pi/4:

sin(3pi/4) - cos(3pi/4) = sqrt(2)/2 - (-sqrt(2)/2)
= 2sqrt(2)/2 = sqrt(2), so it is correct.

Test x = 7pi/4:

sin(7pi/4) - cos(7pi/4) = -sqrt(2)/2 - (sqrt(2)/2) = -2sqrt(2)/2
= -sqrt(2)

Reject x = 7pi/4

Therefore, x = 3pi/4

Answer 4

omg i used to know how to solve these questions
good luck!