# General solution of trig functions

1. Sep 20, 2010

### t_n_p

1. The problem statement, all variables and given/known data

Find the general solution of:
a) sin x = 1/$$\sqrt{2}$$
b) cos x = 0.5

3. The attempt at a solution
a) x = asin(1/$$\sqrt{2}$$)
x = (π/4) + 2nπ

b) x = acos(0.5)
x = π/3 + 2nπ

Basically, my strategy was to solve for the basic angle, and then add multiples of the period (2nπ for sin and cos, nπ for tan).

a) x = 2nπ + asin(1/$$\sqrt{2}$$)
or x = (2n+1)π - asin(1/$$\sqrt{2}$$)

b) 2nπ ± acos(1/2)

Just wondering,

for a) how do they find the second solution x = (2n+1)π - asin(1/$$\sqrt{2}$$)? I'm curious to know, as the follow on question asks for the first 2 positive solutions to sin x = 1/$$\sqrt{2}$$, which can apparently only be obtained using both answers shown above. I only know how to get the first solution

for b) where does the ± come from? My answer is the same but for ±.

2. Sep 20, 2010

### t_n_p

hmm interesting.
got an email notification saying there was a reply to this thread, but there is nothing here!

guess posted must have deleted!

3. Sep 20, 2010

### Chewy0087

Have you sketched both of them and tried to reason by looking at the sketches? It should be quite helpful to do that.

Some things to note; cos is an even function as you'll see when you sketch it which is related to the +-.

Sin is an odd function, and therefore is unlikely to have just one solution for all, hopefully when you sketch it you'll see that.

4. Sep 20, 2010

### t_n_p

I understand graphically how the solutions are valid, but is there a set method/recipe to obtain these results?

Would it be right ro assume, that for the cos question, my method is correct, it just needs the plus/minus in there?

as for the sin question, how they get the second general solution is still a puzzle to me...

5. Sep 20, 2010

### HallsofIvy

Just adding multiples of 2pi is not enough (saying that sin(x) has period 2pi means that if y and x differ by a multiple of 2pi, then sin(x)= sin(y). It does NOT say that if sin(x)= sin(y) then they must differ by a multiple of 2pi.

One way to remember the connections is to think of sine and cosine in terms of the unit circle. If (x, y) is a point on the unit circle so that the radius make angle $\theta$ with the positive x-axis, then $x= cos(\theta)$ and $y= sin(\theta)$. That means that 2 angles, $\theta$ and $\phi$, between 0 and $2\pi$ will have the same sine if they lie on same horizontal line (same y value) and such a line will intersect the unit circle twice: by symmetry, $\phi= \pi- \theta$. Similarly with cosine and a vertical line: $cos(\theta)= cos(2\pi- \theta)$.

6. Sep 20, 2010

### t_n_p

ok yep, definately see what you are saying (halls of ivy).

when I look at sin, positive in 1st and 2nd quadrants, then when I look at the solutions it makes sense. same with cos, positive in 1st and 4th quadrants, the plus minus accounts for this.

Naturally this leads to the following question:

find general solution of:
tanx = -3/$$\sqrt{3}$$
x = atan(-3/$$\sqrt{3}$$)

tan negative in 2nd and fourth quadrants, so general equation(s) are
(2n-1)π - atan(-3/$$\sqrt{3}$$)
2nπ - atan(-3/$$\sqrt{3}$$)

Sound good?

7. Sep 20, 2010

### t_n_p

bump.

the negative in the atan(.....) is bothering me.
I still need to be convinced!

8. Sep 23, 2010

### t_n_p

bump
still seeking clarification with the negative case