# Thread: trig question

1. ## trig question

need help with these questions -

solve sin(2x-45) = √ 3 / 2

and

find general solutions of sin2x=cosx

2. ## Re: trig question

Originally Posted by jshine96
need help with these questions -

solve sin(2x-45) = √ 3 / 2

and

find general solutions of sin2x=cosx

\displaystyle \displaystyle \begin{align*} \sin{(2x - 45)} &= \frac{\sqrt{3}}{2} \\ 2x - 45 &= \left\{ 60, 120 \right\} + 360n , n \in \mathbf{Z} \end{align*}

Go from here.

\displaystyle \displaystyle \begin{align*} \sin{2x} &= \cos{x} \\ 2\sin{x}\cos{x} &= \cos{x} \\ 2\sin{x}\cos{x} - \cos{x} &= 0 \\ \cos{x}\left(2\sin{x} - 1\right) &= 0 \\ \cos{x} = 0 \textrm{ or } 2\sin{x} - 1 &= 0 \end{align*}

Go from here.

3. ## Re: trig question

Originally Posted by jshine96
need help with these questions -

solve sin(2x-45) = √ 3 / 2

and

find general solutions of sin2x=cosx

next time, please post trig questions in the trig forum.

4. ## Re: trig question

sorry, i forgot to add -

solve sin(2x-45) = √ 3 / 2 for -180 ≤ x ≤ 180

5. ## Re: trig question

Originally Posted by jshine96
sorry, i forgot to add -

solve sin(2x-45) = √ 3 / 2 for -180 ≤ x ≤ 180
It's solved in exactly the same way, you just let n be whatever values will give you solutions in the given domain.

6. ## Re: trig question

sorry.. i still don't get how to solve it :L [i get the first part]

7. ## Re: trig question

and for this question find general solutions of sin2x=cosx

i understand what you wrote, but confused on what to do next.. do I solve what x is? like 90, 27, 30, 150?