# Solve cos θ+sin θ=cos 2θ+sin 2θ?

Jun 20, 2018

$\theta \in \left\{2 k \pi\right\} \cup \left\{\frac{2}{3} k \pi + \frac{\pi}{6}\right\} , k \in \mathbb{Z}$.

#### Explanation:

$\cos \theta + \sin \theta = \cos 2 \theta + \sin 2 \theta$.

$\therefore \underline{\cos 2 \theta - \cos \theta} + \underline{\sin 2 \theta - \sin \theta} = 0$.

$\therefore - 2 \sin \left(\frac{3}{2} \theta\right) \sin \left(\frac{1}{2} \theta\right) + 2 \cos \left(\frac{3}{2} \theta\right) \sin \left(\frac{1}{2} \theta\right) = 0$.

$\therefore 2 \sin \left(\frac{1}{2} \theta\right) \left\{\cos \left(\frac{3}{2} \theta\right) - \sin \left(\frac{3}{2} \theta\right)\right\} = 0$.

$\therefore \sin \left(\frac{1}{2} \theta\right) = 0 , \mathmr{and} , \sin \left(\frac{3}{2} \theta\right) = \cos \left(\frac{3}{2} \theta\right)$.

$\text{If, "sin(1/2theta)=0," then, } \frac{1}{2} \theta = k \pi , k \in \mathbb{Z} ,$

$\mathmr{and} , \theta = 2 k \pi , k \in \mathbb{Z}$.

$\text{In case, } \sin \left(\frac{3}{2} \theta\right) = \cos \left(\frac{3}{2} \theta\right) \ldots \ldots \ldots \ldots \left(\ast\right) ,$

$\text{ then, "cos(3/2theta)!=0, because if } \cos \left(\frac{3}{2} \theta\right) = 0 ,$

$\text{ then "(ast) rArr sin(3/2theta)=0," and this contradicts }$

${\sin}^{2} \left(\frac{3}{2} \theta\right) + {\cos}^{2} \left(\frac{3}{2} \theta\right) = 0$.

$\text{So, dividing "(ast)" by "cos(3/2theta)," we get,} \tan \left(\frac{3}{2} \theta\right) = 1$.

$\therefore \frac{3}{2} \theta = k \pi + \frac{\pi}{4} , i . e . , \theta = \frac{2}{3} k \pi + \frac{\pi}{6} , k \in \mathbb{Z}$.

Altogether, $\theta \in \left\{2 k \pi\right\} \cup \left\{\frac{2}{3} k \pi + \frac{\pi}{6}\right\} , k \in \mathbb{Z}$.

Jun 21, 2018

$t = \frac{\pi}{6} + \frac{4 k \pi}{3}$
$t = \frac{5 \pi}{6} + \frac{4 k \pi}{3}$
$t = 4 k \pi$
$t = \left(k + 1\right) 2 \pi$

#### Explanation:

sin 2t + cos 2t = sin t + cos t
Use trig identity; $\sin a + \cos a = \sqrt{2} \sin \left(a + \frac{\pi}{4}\right)$.
In this case:
$\sqrt{2} \sin \left(2 t + \frac{\pi}{4}\right) - \sqrt{2} \sin \left(t + \frac{\pi}{4}\right) = 0$
$\sin \left(2 t + \frac{\pi}{4}\right) - \sin \left(t + \frac{\pi}{4}\right) = 0$ (1)
Use trig identity:
$\sin a - \sin b = 2 \cos \left(\frac{a + b}{2}\right) \sin \left(\frac{a - b}{2}\right)$
$\frac{a + b}{2} = \frac{3 t}{2} + \frac{\pi}{4}$
$\frac{a - b}{2} = \frac{t}{2}$
Equation (1) becomes:
$- 2 \cos \left(\frac{3 t}{2} + \frac{\pi}{4}\right) . \sin \left(\frac{t}{2}\right) = 0$
Either factor should be zero.
a. $\cos \left(\frac{3 t}{2} + \frac{\pi}{4}\right) = 0$
1. $\frac{3 t}{2} + \frac{\pi}{4} = \frac{\pi}{2}$ --> $\frac{3 t}{2} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} + 2 k \pi$
$3 t = \frac{\pi}{2} + 4 k \pi$ --> $t = \frac{\pi}{6} + \frac{4 k \pi}{3}$
2. $\frac{3 t}{2} + \frac{\pi}{4} = \frac{3 \pi}{2}$ --> $\frac{3 t}{2} = \frac{5 \pi}{4} + 2 k \pi$
$3 t = \frac{5 \pi}{2} + 4 k \pi$ --> $t = \frac{5 \pi}{6} + \frac{4 k \pi}{3}$
b. $\sin \left(\frac{t}{2}\right) = 0$
1. $\frac{t}{2} = 2 k \pi$ --> $t = 4 k \pi$
2. $\frac{t}{2} = \pi$ --> $t = 2 \pi + 2 k \pi = \left(k + 1\right) 2 \pi$