1. sin(x) + sqrt2 = - sinx
2. 3 tan(x/2)+3=0
3. solve for cosx ; cos x =1/2
Hello william
1
Add $\displaystyle \sin x$ to both sides, and subtract $\displaystyle \sqrt{2}$ from both sides:
$\displaystyle 2\sin x = - \sqrt{2}$
$\displaystyle \implies \sin x = -\frac{\sqrt{2}}{2}$
Now I'm not sure what sort of answer is needed now. Do you use a calculator to work out a single answer for $\displaystyle x$, giving your answer in degrees? Or do you want lots of answers in radians, involving things like $\displaystyle \frac{\pi}{4}$ and $\displaystyle 2n\pi$?
I don't know. But in degrees, possible answers include $\displaystyle -45^o$, $\displaystyle 225^o$, and so on.
2
$\displaystyle 3 \tan \left(\frac{x}{2}\right) + 3 = 0$
$\displaystyle \implies 3 \tan \left(\frac{x}{2}\right)=-3$
$\displaystyle \implies \tan \left(\frac{x}{2}\right)=-1$
Again, you have lots of possibilities now. One answer is that
$\displaystyle \frac{x}{2} = -45^o$
So $\displaystyle x = -90^o$
There are lots of other possibilities...
3
By now, I'm guessing that you want all the possible values of $\displaystyle x$ if $\displaystyle \cos x = \frac{1}{2}$
I reckon the easiest way is to look at a sketch graph of $\displaystyle y = \cos x$, and work them out from that.
The first positive value is $\displaystyle 60^o$ (or $\displaystyle \frac{\pi}{3}$ radians), but there's another on the other side of the y-axis, where $\displaystyle x = -60^o$ ($\displaystyle -\frac{\pi}{3}$)
The graph repeats every $\displaystyle 360^o$, or $\displaystyle 2\pi$ radians, so you get more answers by adding $\displaystyle 360^o$ ($\displaystyle 2\pi$ radians) on to these two answers. So in degrees:
$\displaystyle -60^o, 60^o, 300^o, 360^o, 660^o, 780^o, ...$
I'll leave it to you to find these answers in radians, if you need to.
Grandad