How to solve 1+2cos^2(x+pi/6)=3sin(pi/3-x)?

Jun 12, 2018

$x = \frac{\pi}{6} + 2 k \pi$
$x = \frac{5 \pi}{6} + 2 k \pi$
$x = \frac{3 \pi}{2} + 2 k \pi$

Explanation:

$1 + {\cos}^{2} \left(x + \frac{\pi}{6}\right) = 3 \sin \left(\frac{\pi}{3} - x\right) \left(1\right)$
Note.
3sin (pi/3 - x) = 3cos (pi/2 - (pi/3 - x) = 3cos (x + pi/6).
Call $\left(x + \frac{\pi}{6}\right) = t$ . The equation (1) becomes:
$2 {\cos}^{2} t - 3 \cos t + 1 = 0$.
Solve this quadratic equation for cos t.
Since a + b + c = 0, use shortcut. The 2 real roots are:
$\cos t = 1$ and $\cos t = \frac{c}{a} = \frac{1}{2}$

a. cos t = cos (x + pi/6) = 1
$x + \frac{\pi}{6} = 2 \pi$
$x = 2 \pi - \frac{\pi}{6} = \frac{5 \pi}{6} + 2 k \pi$
b. $\cos \left(x + \frac{\pi}{6}\right) = \frac{1}{2}$
$x + \frac{\pi}{6} = \pm \frac{\pi}{3}$
1. $x + \frac{\pi}{6} = \frac{\pi}{3}$ --> $x = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} + 2 k \pi$
2. $x + \frac{\pi}{6} = \frac{5 \pi}{3}$ (co-terminal to $x = - \frac{\pi}{3}$)
$x = \frac{5 \pi}{3} - \frac{\pi}{6} = \frac{9 \pi}{6} = \frac{3 \pi}{2} + 2 k \pi$

Jun 12, 2018

$x = \left\{2 k \pi \pm \frac{\pi}{6} , k \in \mathbb{Z}\right\} \cup \left\{2 k \pi - \frac{\pi}{2} , k \in \mathbb{Z}\right\}$

Explanation:

We know that,

color(violet)((1)sintheta=cos(pi/2-theta)

color(blue)((2)costheta=1=cos0=>theta=2kpi+-0=2kpi, kinZZ

color(red)((3)costheta=cosalpha=>theta=2kpi+-alpha,kinZZ

Here,

1+2cos^2(x+pi/6)=3color(violet)(sin(pi/3-x)...toApply(1)

=>1+2cos^2(x+pi/6)=3color(violet)(cos[pi/2-(pi/3-x)]

$\implies 1 + 2 {\cos}^{2} \left(x + \frac{\pi}{6}\right) = 3 \cos \left[\frac{\pi}{2} - \frac{\pi}{3} + x\right]$

$\implies 1 + 2 {\cos}^{2} \left(x + \frac{\pi}{6}\right) = 3 \cos \left(\frac{\pi}{6} + x\right)$

$\implies 2 {\cos}^{2} \left(x + \frac{\pi}{6}\right) - 3 \cos \left(x + \frac{\pi}{6}\right) + 1 = 0$

Let , $\cos \left(x + \frac{\pi}{6}\right) = m$

$\therefore 2 {m}^{2} - 3 m + 1 = 0$

$\implies 2 {m}^{2} - 2 m - m + 1 = 0$

$\implies 2 m \left(m - 1\right) - 1 \left(m - 1\right) = 0$

$\implies \left(m - 1\right) \left(2 m - 1\right) = 0$

$\implies m - 1 = 0 \mathmr{and} 2 m - 1 = 0$

$\implies m = 1 \mathmr{and} m = \frac{1}{2} , w h e r e , m = \cos \left(x + \frac{\pi}{6}\right)$

$\implies \cos \left(x + \frac{\pi}{6}\right) = 1 \mathmr{and} \cos \left(x + \frac{\pi}{6}\right) = \frac{1}{2}$

$\left(i\right) \cos \left(x + \frac{\pi}{6}\right) = 1$

=>color(blue)(x+pi/6=2kpi, kinZZ...toApply(2)

$\implies x = 2 k \pi - \frac{\pi}{6} , k \in \mathbb{Z}$

$\left(i i\right) \cos \left(x + \frac{\pi}{6}\right) = \frac{1}{2} = \cos \left(\frac{\pi}{3}\right)$

=>color(red)(x+pi/6=2kpi+-pi/3.kinZZ...toApply(3)

$\implies x + \frac{\pi}{6} = 2 k \pi + \frac{\pi}{3} \mathmr{and} x + \frac{\pi}{6} = 2 k \pi - \frac{\pi}{3} , k \in \mathbb{Z}$

$\implies x = 2 k \pi + \frac{\pi}{3} - \frac{\pi}{6} \mathmr{and} x = 2 k \pi - \frac{\pi}{3} - \frac{\pi}{6} , k \in \mathbb{Z}$

$\implies x = 2 k \pi + \frac{\pi}{6} , \mathmr{and} x = 2 k \pi - \frac{\pi}{2} , k \in \mathbb{Z}$

Hence,

$x = 2 k \pi - \frac{\pi}{6} \mathmr{and} x = 2 k \pi + \frac{\pi}{6} \mathmr{and} x = 2 k \pi - \frac{\pi}{2} , k \in \mathbb{Z}$

$i . e . x = \left\{2 k \pi \pm \frac{\pi}{6} , k \in \mathbb{Z}\right\} \cup \left\{2 k \pi - \frac{\pi}{2} , k \in \mathbb{Z}\right\}$