Trigonometric equations[1 MORE]

**Hellbent** · February 9th 2010, 04:16 AM

Hi,

Could someone check this for me please.

Solve

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

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

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

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

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

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

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

$= \frac{8}{12}$

$= \frac{2}{3}$

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

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

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

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

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

$\Rightarrow$ $sin{\theta} > 0$

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

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

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

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

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

**Sudharaka** · February 9th 2010, 05:24 AM

Originally Posted by Hellbent

Hi,

Could someone check this for me please.

Solve

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

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

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

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

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

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

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

$= \frac{8}{12}$

$= \frac{2}{3}$

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

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

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

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

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

$\Rightarrow$ $sin{\theta} > 0$

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

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

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

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

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

Dear Hellbent,

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

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

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

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

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

Now the general solution,

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

**Hellbent** · February 9th 2010, 05:32 AM

Cool thanks

**Sudharaka** · February 9th 2010, 05:51 AM

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

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

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

Hope this will help you.

**Hellbent** · February 9th 2010, 12:43 PM

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

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

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

Hope this will help you.

I got it

Cool, thanks.

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