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

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

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Answer 1 · 2016-07-14T00:33:06

$\frac{π}{4} and \frac{π}{2}$ $pi/4 and pi/2$

Explanation:

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

Answer 2 · 2016-07-14T00:36:56

$2 sin θ cos θ - 1 = 1 - 2 {sin}^{2} θ$ $2sinthetacostheta - 1 = 1 - 2sin^2theta$

$2 sin θ cos θ - 1 - 1 + 2 {sin}^{2} θ = 0$ $2sinthetacostheta - 1 - 1 + 2sin^2theta = 0$

$2 (sin θ cos θ - 1 + {sin}^{2} θ) = 0$ $2(sinthetacostheta - 1 + sin^2theta) = 0$

Since ${sin}^{2} θ - 1 = - {cos}^{2} θ$ $sin^2theta - 1 = -cos^2theta$ :

$sin θ cos θ - {cos}^{2} θ = 0$ $sinthetacostheta - cos^2theta = 0$

$cos θ (sin θ - cos θ) = 0$ $costheta(sin theta - costheta) = 0$

Solving $cos θ = 0$ $costheta = 0$ 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 · 2016-07-14T00:47:48

$sin2theta-1=cos2theta$

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

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

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

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

When $costheta=0$

then $theta=npi+pi/2$

Again when $sintheta-costheta=0$

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

Then $theta=npi+pi/4$

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

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