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

[karush]

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

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

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

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

$\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]

Re: 2costheta+2sintheta=sqrt6

Hello, karush!

You have: .$$\sin{2\theta}\:=\:\tfrac{1}{2}$$

Then: .$$2\theta \:=\:\begin{Bmatrix}\tfrac{\pi}{6} \\ \tfrac{5\pi}{6} \end{Bmatrix}$$

Therefore: .$$\theta \:=\:\begin{Bmatrix}\tfrac{\pi}{12} \\ \tfrac{5\pi}{12}\end{Bmatrix}$$

 

[Prove It]

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]

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]

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]

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]

Re: 2costheta+2sintheta=sqrt6

so how could this done without squaring?

 

[MarkFL]

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:

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

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