1. ## Trigonometric equations[1 MORE]

Hi,

Could someone check this for me please.

Solve

$\displaystyle 6 cos^2{\theta} + sin{\theta} = 4\, for\, 0 \le x \le {\pi}$

$\displaystyle 6(1 - sin^2{\theta}) + sin{\theta} = 4$

$\displaystyle -6sin^2{\theta} + sin{\theta} + 2 = 0$

$\displaystyle \text{Let}$ $\displaystyle sin{\theta} = a, |a| \le 1$

$\displaystyle 6a^2 - a - 2 = 0$

$\displaystyle D = 1 - 4(6)(-2) = 49$

$\displaystyle a_1 = \frac{-(-1) + \sqrt49}{(2)(6)}$

$\displaystyle = \frac{8}{12}$

$\displaystyle = \frac{2}{3}$

$\displaystyle a_2 = \frac{-(-1) - \sqrt49}{(2)(6)}$

$\displaystyle = \frac{-1}{2}$

$\displaystyle sin{\theta}_1 = \frac{2}{3}$

$\displaystyle sin{\theta}_2 = \frac{-1}{2}$

$\displaystyle 0 \le x \le {\pi}$

$\displaystyle \Rightarrow$$\displaystyle sin{\theta} > 0 \displaystyle \Rightarrow \displaystyle sin{\theta} = \frac{2}{3} \displaystyle {\theta} = (-1)^n \displaystyle arcsin \displaystyle \frac{2}{3} \displaystyle + {\pi}n \displaystyle \text{Solution:}\,$$\displaystyle {\theta}$$\displaystyle = arcsin \frac{2}{3}, n = 0 \displaystyle {\theta} = {\pi} - arcsin \frac{2}{3}, n = 1 2. Originally Posted by Hellbent Hi, Could someone check this for me please. Solve \displaystyle 6 cos^2{\theta} + sin{\theta} = 4\, for\, 0 \le x \le {\pi} \displaystyle 6(1 - sin^2{\theta}) + sin{\theta} = 4 \displaystyle -6sin^2{\theta} + sin{\theta} + 2 = 0 \displaystyle \text{Let} \displaystyle sin{\theta} = a, |a| \le 1 \displaystyle 6a^2 - a - 2 = 0 \displaystyle D = 1 - 4(6)(-2) = 49 \displaystyle a_1 = \frac{-(-1) + \sqrt49}{(2)(6)} \displaystyle = \frac{8}{12} \displaystyle = \frac{2}{3} \displaystyle a_2 = \frac{-(-1) - \sqrt49}{(2)(6)} \displaystyle = \frac{-1}{2} \displaystyle sin{\theta}_1 = \frac{2}{3} \displaystyle sin{\theta}_2 = \frac{-1}{2} \displaystyle 0 \le x \le {\pi} \displaystyle \Rightarrow$$\displaystyle sin{\theta} > 0$

$\displaystyle \Rightarrow$ $\displaystyle sin{\theta} = \frac{2}{3}$

$\displaystyle {\theta} = (-1)^n$

$\displaystyle arcsin$ $\displaystyle \frac{2}{3}$ $\displaystyle + {\pi}n$

$\displaystyle \text{Solution:}\,$$\displaystyle {\theta}$$\displaystyle = arcsin \frac{2}{3}, n = 0$

$\displaystyle {\theta} = {\pi} - arcsin \frac{2}{3}, n = 1$
Dear Hellbent,

It would be easier if you factorize the equation other than using the general solution of a quadratic equation.

That is, $\displaystyle -6sin^2{\theta} + sin{\theta} + 2 = 0$

$\displaystyle (3sin\theta-2)(2sin\theta+1)=0$

$\displaystyle Since,~0\leq\theta\leq\pi\Rightarrow{0\leq{Sin}\th eta\leq{1}}$

$\displaystyle Sin\theta=\frac{2}{3}$

Now the general solution,

$\displaystyle \theta=n\pi+(-1)^{n}Sin^{-1}\frac{2}{3}$

3. Cool thanks

4. Originally Posted by Hellbent
Cool thanks
Dear Hellbent,

Sorry I did'nt help you when I first saw your thread. And that must be the reason for you to send me a privet message. That is because I just came home after a tiresome days work and I did'nt have that feeling to answer threads.

Anyway now do you know your mistake,

When, $\displaystyle Sin\theta=\frac{2}{3}$

$\displaystyle \theta=arcsin\frac{2}{3}$ gives all the solutions that could be taken for $\displaystyle \theta$. When writing the general solution you must take $\displaystyle \theta$ such that it is in the first quadrant. That is why I used $\displaystyle Sin^{-1}\frac{2}{3}$. If this seems a bit confusing please refer,

General Solutions of Trigonometric Functions, Maths First, Institute of Fundamental Sciences, Massey University

5. Originally Posted by Sudharaka
Dear Hellbent,

Sorry I did'nt help you when I first saw your thread. And that must be the reason for you to send me a privet message. That is because I just came home after a tiresome days work and I did'nt have that feeling to answer threads.

Anyway now do you know your mistake,

When, $\displaystyle Sin\theta=\frac{2}{3}$

$\displaystyle \theta=arcsin\frac{2}{3}$ gives all the solutions that could be taken for $\displaystyle \theta$. When writing the general solution you must take $\displaystyle \theta$ such that it is in the first quadrant. That is why I used $\displaystyle Sin^{-1}\frac{2}{3}$. If this seems a bit confusing please refer,

General Solutions of Trigonometric Functions, Maths First, Institute of Fundamental Sciences, Massey University