# How do you solve cos 2x − 2 sin2 x = 0 interval (−5, 5]?

Jun 24, 2018

-256^@72; -166^@72; -76^@72; 13^@28; 103^@28; 193^@28

#### Explanation:

cos 2x - 2sin 2 (1)
x = 0
Divide both sides by cos 2x.
Condition: $\cos 2 x \ne 0$, --> $2 x \ne \frac{\pi}{2}$, and $2 x \ne \frac{3 \pi}{2}$, -->
$x \ne \frac{\pi}{4}$, and $x \ne \frac{3 \pi}{4}$.
Equation (1) becomes:
1 - 2 tan 2x = 0
$\tan 2 x = \frac{1}{2}$
Calculator and unit circle give -->
$2 x = {56}^{\circ} 56 + k {180}^{\circ}$.
$x = {13}^{\circ} 28 + k {90}^{\circ}$
The interval (-5, 5) is equivalent to interval $\left(- {286}^{\circ} 62 , {286}^{\circ} 62\right)$
k = 0 --> x = 13.28
k = -1 --> x = 13.28 - 90 = - 76.72 -->
k = -2 --> x = 13.28 - 180 = -16672
k = -3 --> x = 13.28 + 270 = -256.72
k = 1 --> x= 13.28 + 90 = 103.28
k = 2 --> x = 13.28 + 180 = 193.28
The answers for (-5, 5) are:
-256^@72; -166^@72; -76^@72; 13^@28; 103^@28; 193^@28

Jun 24, 2018

$x = \text{Arc""tan} \left(\frac{1}{2} \left(1 - \sqrt{5}\right)\right) \pm {45}^{\circ} + {180}^{\circ} k \quad$ integer $k$

#### Explanation:

I'll take the questioner at their word and solve

$\cos 2 x - 2 \sin 2 x = 0$

Hopefully they didn't mean ${\sin}^{2} x$ or ${\cos}^{2} x$

There are a few different ways I can think of to solve this: (A) Let $y = 2 x$ and invert $\tan y$. (B) Use the double angle identities, then square to get a solvable quadratic for ${\cos}^{2} x .$ (C) The linear combination of sine and cosine of the same angle is a scaling and a phase shift.

Let's try (C) first. We convert $\left(1 , - 2\right)$ to polar coordinates:

$| \left(1 \text{,} - 2\right) | = \sqrt{{1}^{2} + {\left(- 2\right)}^{2}} = \sqrt{5}$

This is fourth quadrant so we can use the principal value of the inverse tangent.

alpha = angle(1,-2) = "Arc""tan"(-2) approx -63.43°

We have

$1 = \sqrt{5} \cos \alpha$

$- 2 = \sqrt{5} \sin \alpha$

$1 \cos 2 x - 2 \sin 2 x = 0$

$\sqrt{5} \cos \alpha \cos 2 x + \sqrt{5} \sin \alpha \sin 2 x = 0$

$\sqrt{5} \cos \left(2 x - \alpha\right) = 0$

$\cos \left(2 x - \alpha\right) = \cos \left({90}^{\circ}\right)$

$2 x - \alpha = \pm {90}^{\circ} + {360}^{\circ} k$

$x = \frac{\alpha}{2} \pm {45}^{\circ} + {180}^{\circ} k$

$x = \frac{1}{2} \text{Arc""tan} \left(- 2\right) \pm {45}^{\circ} + {180}^{\circ} k$

The factor of $\frac{1}{2}$ on the arctangent is awkward.

We can make progress on

$\theta = \frac{1}{2} \alpha$ where $\alpha = \angle \left(1 , - 2\right) = \text{Arc""tan} \left(- 2\right)$

We know $\alpha$'s in the fourth quadrant so

$\cos \alpha = \frac{1}{\sqrt{5}}$

$\sin \alpha = - \frac{2}{\sqrt{5}}$

$\tan \left(\frac{\alpha}{2}\right) = \sin \frac{\alpha}{1 + \cos \alpha} = \frac{1 - \cos \alpha}{\sin} \alpha$

$\tan \left(\frac{\alpha}{2}\right) = \frac{1 - \frac{1}{\sqrt{5}}}{- \frac{2}{\sqrt{5}}} = \frac{1}{2} \left(1 - \sqrt{5}\right)$

$\frac{\alpha}{2} = \frac{1}{2} \text{Arc""tan"(-2) = "Arc""tan} \left(\frac{1}{2} \left(1 - \sqrt{5}\right)\right)$

$x = \text{Arc""tan} \left(\frac{1}{2} \left(1 - \sqrt{5}\right)\right) \pm {45}^{\circ} + {180}^{\circ} k$

I gotta go so I'll let someone else convert to radians, pull out the answers in range and check it.

~~~~~~~~~~~~~

Attempts at other approaches:

Let $y = 2 x$

$\cos y - 2 \sin y = 0$

$\cos y = 2 \sin y$

$\tan y = \sin \frac{y}{\cos} y = \frac{1}{2}$

Invoking the multivalued arctangent,

$y = \arctan \left(\frac{1}{2}\right)$

$2 x = \arctan \left(\frac{1}{2}\right)$

$x = \frac{1}{2} \arctan \left(\frac{1}{2}\right)$

Similar to where we were but simpler.

~~~~~~~~~~~~~~~~~~

$2 {\cos}^{2} x - 1 - 2 \cos x \sin x = 0$

$2 {\cos}^{2} x - 1 = 2 \cos x \sin x$

Let's square to get square of sines and cosines.

${\left(2 {\cos}^{2} x - 1\right)}^{2} = 4 {\cos}^{2} x {\sin}^{2} x = 4 {\cos}^{2} x \left(1 - {\cos}^{2} x\right)$

Let $y = {\cos}^{2} x$

${\left(2 y - 1\right)}^{2} = 4 y \left(1 - y\right)$

$4 {y}^{2} - 4 y + 1 = 4 y - 4 {y}^{2}$

$8 {y}^{2} - 8 y + 1 = 0$

$y = \frac{1}{8} \left(4 \pm \sqrt{8}\right) = \frac{1}{4} \left(2 \pm \sqrt{2}\right)$

Uncle.