# Arctrig and Solving Trig. Equation

1. Aug 24, 2009

### cyspope

1. The problem statement, all variables and given/known data
I have to find out the radian on the unit circle based on the questions below.
1. tan3x=1
2. 2sin(2t)+1=0

2. Relevant equations

3. The attempt at a solution
1. tan3x=1
tanx= (1/3) ?

2. 2sin(2t)+1=0
2sin(2t)= -1
sin2t= (-1/2) ?

2. Aug 24, 2009

### rock.freak667

1) No (do what you did in question 2)

2) Yes.

continue on now

3. Aug 24, 2009

### cyspope

I didn't really know what to do after that.
Could you give me a hint.

4. Aug 24, 2009

### rock.freak667

if tanA=x, then A=tan-1(x). This is similar for sin and cos.

5. Aug 24, 2009

### cyspope

Thank you so much. It did help me a lot.

6. Aug 24, 2009

### VietDao29

If you want to find the general solution, then you may want to memorize these formulae:

$$\sin(x) = y \Leftrightarrow \left[ \begin{array}{ccc} x & = & \arcsin(y) + 2k \pi \\ x & = & \pi - \arcsin(y) + 2k' \pi \end{array} \right. , k , k' \in \mathbb{Z}$$

Since sin has the period of $$2 \pi$$, and $$\sin(\pi - x) = \sin(x)$$.

$$\cos(x) = y \Leftrightarrow \pm \arccos(y) + 2k \pi , k \in \mathbb{Z}$$

Since cos has the period of $$2 \pi$$, and $$\cos(- x) = \cos(x)$$.

$$\tan(x) = y \Leftrightarrow \arctan(y) + k \pi , k \in \mathbb{Z}$$

Since tan has the period of $$\pi$$.

$$\cot(x) = y \Leftrightarrow \mbox{arccot}(y) + k \pi , k \in \mathbb{Z}$$

Since cot has the period of $$\pi$$.