How do you solve 5sinx+2=sinx for 0<=x<=2pi?

Nov 20, 2016

$x \in \left\{\frac{\pi}{6} , \frac{11 \pi}{6}\right\}$

Explanation:

We start as if this wasn't a trigonometric equation:
$5 \sin x + 2 = \sin x$ For now we'll 'pretend' that its domain is ${D}_{0} = \setminus m a t h \boldsymbol{R}$ (all $x$s that make sense) and at the end we'll adjust the solutions to its actual domain $D = \left[0 , 2 \pi\right]$.

$5 \sin x + 2 = \sin x$
$5 \sin x - \sin x = - 2$
$4 \sin x = - 2$
$\sin x = - \frac{2}{4}$
$\sin x = - \frac{1}{2}$

Now, here we have at least two ways to go, I personally would use the fact that sine is an odd function:

$- \sin x = \frac{1}{2}$
$\sin \left(- x\right) = \frac{1}{2}$

$- x = \frac{\pi}{6} + 2 k \pi \mathmr{and} - x = \pi - \frac{\pi}{6} + 2 k \pi , k \in m a t h \boldsymbol{Z}$
$x = - \frac{\pi}{6} + 2 k \pi \mathmr{and} x = - \frac{5 \pi}{6} + 2 k \pi , k \in m a t h \boldsymbol{Z}$
(Note: during the last step we multiplied both equations by $\left(- 1\right)$ and yet $+ 2 k \pi$ stayed $+ 2 k \pi$ - didn't change to $- 2 k \pi$. The reason for that it that here it doesn't really matter because $k$ goes trough the set of all integers - both positive and negative. I'm just used to writing $+ s o m e \pi$)

Now we can adjust our solution to the domain $D = \left[0 , 2 \pi\right]$. Concluding:
$x \in \left\{\frac{7 \pi}{6} , \frac{11 \pi}{6}\right\}$