solve sin2x - 1 = cos2x where x is between 0 and 360

[hndalama]

Solve sin2x - 1 = cos2x, where x is between 0 and 360 degrees

My attempt went as follows:

sin2x = cos2x +1
sin[SUP]2[/SUP]2x = (cos2x + 1)[SUP]2[/SUP]
1 - cos[SUP]2[/SUP]2x = cos[SUP]2[/SUP]2x + 2cos2x + 1
0 = 2cos[SUP]2[/SUP]2x + 2cos2x
0 = (2cos2x)(cos2x + 1)

2cos2x = 0  , 2x = 90, 270, 450, 630  , x = 45, 135, 225, 315

cos2x = -1   , 2x = 180, 540   , x = 90, 270

the answer is x = 90, 270, 45, 225

Why is 135 and 315 not correct? Is there a better way of solving this equation?

 

[stapel]

Solve sin2x - 1 = cos2x, where x is between 0 and 360 degrees

My attempt went as follows:

sin2x = cos2x +1
sin²2x = (cos2x + 1)²

Squaring both sides is what is called "an irreversible step", and it often introduces new solutions which did not exist for the original equation. For instance:

. . . . .\(\displaystyle x\, =\, 4\)

This equation has one solution, being x = 4. But this equation:

. . . . .\(\displaystyle x^2\, =\, 16\)

...has two solutions, being x = -4 and x = +4. Only one of these solutions is valid for the original equation. This is why one must check all solutions when one has squared to obtain those solutions.

1 - cos²2x = cos²2x + 2cos2x + 1
0 = 2cos²2x + 2cos2x
0 = (2cos2x)(cos2x + 1)

2cos2x = 0 , 2x = 90, 270, 450, 630 , x = 45, 135, 225, 315

cos2x = -1 , 2x = 180, 540 , x = 90, 270

the answer is x = 90, 270, 45, 225

Why is 135 and 315 not correct?

Because of the squaring of both sides.

Is there a better way of solving this equation?

Try using other identities. You know that:

. . . . .\(\displaystyle \sin(2x)\, =\, 2\, \sin(x)\, \cos(x)\)

You also know that:

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

What happens if you apply these identities?

. . . . .\(\displaystyle \sin(2x)\, =\, \cos(2x)\, +\, 1\)

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

Where does this lead?

 

[hndalama]

. . . . .\(\displaystyle \sin(2x)\, =\, \cos(2x)\, +\, 1\)

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

Where does this lead?

I could only simplify up to
sinxcosx = cos²x
from here I don't know the next step
how do i convert this to an equation with the same trig function?

 

[stapel]

I could only simplify up to
sinxcosx = cos²x
from here I don't know the next step
how do i convert this to an equation with the same trig function?

What happens if you do the customary step of putting everything on one side of the "equals" sign, and factoring?

 

[hndalama]

What happens if you do the customary step of putting everything on one side of the "equals" sign, and factoring?

cosx(cosx - sinx) = 0

cosx = 0 x = 90, 270

cosx - sinx = 0
1 - tanx = 0
tanx = 1 x=45, 225

thank you