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Trigonometry Trigonometric equation: 2cos(θ) + 2sin(θ) = √(6)

karush

Well-known member
Jan 31, 2012
2,714
$2\cos{\theta}+2\sin{\theta}=\sqrt{6}$

$\displaystyle\cos{\theta}+\sin{\theta}=\frac{ \sqrt{6} }{2}$

$\displaystyle(\cos{\theta}+\sin{\theta})^2=\frac{3}{2}$

$\displaystyle\cos^2{\theta}+2cos\theta\sin\theta+\sin{\theta}^2=\frac{3}{2}$

$\displaystyle\sin{2\theta}=\frac{1}{2}
\Rightarrow
\theta=\frac{\pi}{12}=15^o
$

the other answer is $75^o$ but don't know how you get it.
 

soroban

Well-known member
Feb 2, 2012
409
Re: 2costheta+2sintheta=sqrt6

Hello, karush!

You have: .[tex]\sin{2\theta}\:=\:\tfrac{1}{2}[/tex]

Then: .[tex]2\theta \:=\:\begin{Bmatrix}\tfrac{\pi}{6} \\ \tfrac{5\pi}{6} \end{Bmatrix}[/tex]

Therefore: .[tex]\theta \:=\:\begin{Bmatrix}\tfrac{\pi}{12} \\ \tfrac{5\pi}{12}\end{Bmatrix}[/tex]
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: 2costheta+2sintheta=sqrt6

Since you have gotten results through squaring, you should also check to see if any of the results are extraneous. Also, even though you have gotten the solutions from one cycle, there are not all the solutions.
 

karush

Well-known member
Jan 31, 2012
2,714
Re: 2costheta+2sintheta=sqrt6

I was lazy I looked at graph on W|A and saw only 2 solutions.
Actually not how to seek find more posibilities
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: 2costheta+2sintheta=sqrt6

$2\cos{\theta}+2\sin{\theta}=\sqrt{6}$

$\displaystyle\cos{\theta}+\sin{\theta}=\frac{ \sqrt{6} }{2}$

$\displaystyle(\cos{\theta}+\sin{\theta})^2=\frac{3}{2}$

$\displaystyle\cos^2{\theta}+2cos\theta\sin\theta+\sin{\theta}^2=\frac{3}{2}$

$\displaystyle\sin{2\theta}=\frac{1}{2}
\Rightarrow
\theta=\frac{\pi}{12}=15^o
$

the other answer is $75^o$ but don't know how you get it.
To get the other solution, consider the identity:

\(\displaystyle \sin(\pi-x)=\sin(x)\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Re: 2costheta+2sintheta=sqrt6

There are TWO points on the unit circle where $y = \sin \phi = \dfrac{1}{2}$.

One is the point $(\frac{\sqrt{3}}{2},\frac{1}{2})$ and one is the point $(\frac{-\sqrt{3}}{2},\frac{1}{2})$.

The first corresponds to the value you found:

$\phi = 2\theta = \frac{\pi}{6} \implies \theta = \frac{\pi}{12} = 15^{\circ}$

The second corresponds to:

$\phi = 2\theta = \frac{5\pi}{6} \implies \theta = \frac{5\pi}{12} = 75^{\circ}$

That said, I'm surprised no one has mentioned the cases:

$\theta = \dfrac{13\pi}{12}$

$\theta = \dfrac{17\pi}{12}$

which also satisfy the equation:

$\sin 2\theta = \dfrac{1}{2}$

but which do not satisfy the original equation (these are the "extraneous" solutions introduced by the squaring).

Finally, (to give an example) what happens if:

$\theta = \dfrac{29\pi}{12}$...hmm?
 

karush

Well-known member
Jan 31, 2012
2,714
Re: 2costheta+2sintheta=sqrt6

so how could this done without squaring?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: 2costheta+2sintheta=sqrt6

so how could this done without squaring?
You could use a linear combination identity:

\(\displaystyle \cos(\theta)+\sin(\theta)=\frac{\sqrt{6}}{2}\)

\(\displaystyle \sin\left(\theta+\frac{\pi}{4} \right)=\frac{\sqrt{3}}{2}\)

Quadrant I solution:

\(\displaystyle \theta+\frac{\pi}{4}=\frac{\pi}{3}\)

\(\displaystyle \theta=\frac{\pi}{12}\)

Quadrant II solution:

\(\displaystyle \theta+\frac{\pi}{4}=\frac{2\pi}{3}\)

\(\displaystyle \theta=\frac{5\pi}{12}\)