Question

How do you solve `sin 2theta -1 = cos 2 theta`?

Answer 1

`pi/4 and pi/2`

Explanation:

Re-write the equation:
sin 2t - cos 2t = 1
Use the trig identity:
`sin a - cos a = sqrt2sin (a - pi/4).`
Replace a by 2t -->
`sin 2t - cos 2t = sqrt2 sin (2t - pi/4) = 1`
`sin (2t - pi/4) = 1/sqrt2 = sqrt2/2`
Trig table and unit circle -->
`sqrt2/2` is `sin ((pi)/4)` and in the same time sin `((3pi)/4)`. There for:
a. `(2t - pi/4) = pi/4`
`2t = pi/4 + pi/4 = pi/2` --> `t = pi/4`
b. `(2t - pi/4) = (3pi)/4`
`2t = (3pi)/4 + pi/4 = pi` --> `t = pi/2`
Check
a. `t = pi/4` --> `2t = pi/2` --> sin 2t = 1 and cos 2t = 0
We have 1 = 1 . OK
b. `t = pi/2` --> `2t = pi` --> sin 2t = 0, and -cos 2t = -(-1) = 1.
We have 1 = 1. OK

Answer 2

`2sinthetacostheta - 1 = 1 - 2sin^2theta`

`2sinthetacostheta - 1 - 1 + 2sin^2theta = 0`

`2(sinthetacostheta - 1 + sin^2theta) = 0`

Since `sin^2theta - 1 = -cos^2theta`:

`sinthetacostheta - cos^2theta = 0`

`costheta(sin theta - costheta) = 0`

Solving `costheta = 0` first:

`theta = 90˚, 270˚`

Solving `sintheta - costheta = 0`:

`sqrt((sin theta - costheta)^2) = 0`

`sqrt(sin^2theta - 2sinthetacostheta + cos^2theta) = 0`

`(sqrt(sin^2theta - 2sinthetacostheta + cos^2theta))^2 = 0^2`

`sin^2theta + cos^2theta - 2sinthetacostheta = 0`

Since `sin^2theta + cos^2theta = 1` and `2sinthetacostheta = sin2theta`:

`1 - sin2theta = 0`

`1 = sin2theta`

`theta = 45˚`

Hence, the solutions are `theta = 45˚, 90˚ and 270˚`.

Hopefully this helps!

Answer 3

`sin2theta-1=cos2theta`

`=>2sinthetacostheta-(1+cos2theta)=0`

`=>2sinthetacostheta-2cos^2theta=0`

`=>2costheta(sintheta-costheta)=0`

`:.costheta=0 and sintheta-costheta=0`

When `costheta=0`

then `theta=npi+pi/2`

Again when `sintheta-costheta=0`

`=>tantheta=1=tan(pi/4)`

Then `theta=npi+pi/4`

where `n=0,+-1,+-2,+-3...`
.