Solving trigonometry equations

[fordy2707]

Hi all, can you show me how to calculate this and breakdown how you get to the answer for me to Understand ,I have been shown what is believed to be the answer from the notes but not a clue How it was reached.

Find all the solutions to the following equation at interval 0,6PI

cos(x) + 2cos^2(x)=0

=

Cosx(1+2cosx)=0
Cosx=cos(pie/2)
X=2npie+_pie/2
Cosx=cos(2pie/3)

Many thanks

 

[kaliprasad]

Hi all, can you show me how to calculate this and breakdown how you get to the answer for me to Understand ,I have been shown what is believed to be the answer from the notes but not a clue How it was reached.

Find all the solutions to the following equation at interval 0,6PI

cos(x) + 2cos^2(x)=0

=

Cosx(1+2cosx)=0
Cosx=cos(pie/2)
X=2npie+_pie/2
Cosx=cos(2pie/3)

Many thanks

your approach is good. 1st you need to find a particular solution then from it all solutions in the given range.

$\cos(x)(1+2\cos(x)) = 0$
gives $\cos(x) = 0$ and we have particular solution $x = \dfrac{\pi}{2}$ givinin solution in the range
$x = \dfrac{\pi}{2}+ n \pi $ ( n is from 0 to 5) and ,$x =- \dfrac{\pi}{2}+ n \pi $ ( n is from 1 to 6) so that value is in the range

2nd set of solution
$\cos(x) = \frac{-1}{2}$ or $ x= \frac{2\pi}{3}$ particular solution

$x = \dfrac{2\pi}{3}+ n \pi $ ( n is from 0 to 5) and ,$x =- \dfrac{2\pi}{3}+ n \pi $ ( n is from 1 to 6) so that value is in the range

 

[fordy2707]

0-6PI is that the equivalent of 3 full cycles ,1080 degress ?

 

[topsquark]

Cosx(1+2cosx)=0
Cosx=cos(pie/2)
X=2npie+_pie/2
Cosx=cos(2pie/3)

(Shake) \(\displaystyle \pi\) is spelled "pi." Pie refers to something you eat for breakfast. (Sun)

-Dan

 

[soroban]

Hi all, can you show me how to calculate this
and breakdown how you get to the answer for me to Understand.
I have been shown what is believed to be the answer from the notes but not a clue how it was reached.

Find all the solutions to the following equation at interval $$(0,\,6\pi).$$

. . $$\cos(x) + 2\cos^2(x) \;=\;0$$

You have a quadratic equation.

Factor: .$$\cos x(1 + 2\cos x) \;=\;0$$

Set each factor equal to zero and solve.

. . $$\cos x \;=\;0 \quad\Rightarrow\quad x \;=\;\tfrac{\pi}{2} + \pi n$$

. . $$1 + 2\cos x \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\tfrac{1}{2} \quad\Rightarrow\quad x \:=\:\left( \begin{array}{cc}\frac{2\pi}{3} + 2\pi n \\ \frac{4\pi}{3} + 2\pi n \end{array}\right)$$