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

#### karush

##### Well-known member
$2\cos{\theta}+2\sin{\theta}=\sqrt{6}$

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

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

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

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

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

#### soroban

##### Well-known member
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

##### Well-known member
MHB Math Helper
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
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

Staff member
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
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
Re: 2costheta+2sintheta=sqrt6

so how could this done without squaring?

#### MarkFL

Staff member
Re: 2costheta+2sintheta=sqrt6

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

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

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

$$\displaystyle \theta+\frac{\pi}{4}=\frac{\pi}{3}$$
$$\displaystyle \theta=\frac{\pi}{12}$$
$$\displaystyle \theta+\frac{\pi}{4}=\frac{2\pi}{3}$$
$$\displaystyle \theta=\frac{5\pi}{12}$$