solving trig problems using additional formulas

oh ok so basically to find the next solution you need to add 180 to the first solution, does this always make sense for tan questions tho?

Yep.

tan(x - 180^o) = tan(x) = tan(x + 180^o)... and so forth.

but why for the question which i attached you here the answer has been added two times once by pi and once by 2pi?

Presumably the calculation gives you a negative answer then? For a tan(x) question, there will always be two solutions in the range (0<x<2pi).

that's right i had a negative answer of -0.296 which has been added by pi to give me 2.84 and by 2pi to give me 5.99.
do you suggest that overtime i get a negative answer for tan x i should add the answer once by pi and once by 2pi?

Yeah, like I say, you're looking for all the possible solutions in that range - so add any multiples of pi that'll get you those solutions.

can't add by 3pi this because of the range is that right?

Usually adding 3pi to the answer you get would take it above 2pi, so yeah.

appreciate your help mate

appreciate your help mate

No problem

oh forgot to ask does this rule apply for cos and sin as well?

$\tan$ is periodic with period $\pi$, so if $x$ is a solution to $\tan x = k$, then so is $x + \pi$.

$\sin$ is periodic with period $2\pi$, so if $x$ is a solution to $\sin x = k_s$ or $\cos x = k_c$, then so is $x+2\pi$.

...

does the sin periodic will be the same as cos periodic?

Yes, I forgot to say that sorry. $\sin$ and $\cos$ have the same period. They are just shifts of each other, afterall.