finding solutions in trig

**NoAsherelol** · June 30th 2008, 04:49 PM

Hey i need help with this problem

sin 3x= sqrt3/2

**Jonboy** · June 30th 2008, 04:52 PM

You mean this, right?

$Sin\,3x = \frac{\sqrt{3}}{2}$

**mr fantastic** · June 30th 2008, 04:54 PM

Originally Posted by NoAsherelol

Hey i need help with this problem

sin 3x= sqrt3/2

Where are you stuck? What have you tried?

**NoAsherelol** · June 30th 2008, 04:56 PM

Originally Posted by Jonboy

You mean this, right?

$Sin\,3x = \frac{\sqrt{3}}{2}$

i meant $Sin\,2x = \frac{\sqrt{3}}{2}$

**mr fantastic** · June 30th 2008, 05:00 PM

Originally Posted by NoAsherelol

i meant $Sin\,2x = \frac{\sqrt{3}}{2}$

The questions asked in post # 3 still apply.

**ticbol** · June 30th 2008, 05:02 PM

Originally Posted by NoAsherelol

i meant $Sin\,2x = \frac{\sqrt{3}}{2}$

Man, this made me really laugh!

**NoAsherelol** · June 30th 2008, 05:03 PM

Originally Posted by mr fantastic

The questions asked in post # 3 still apply.

ih im sorry, this is my problem, $Sin\,x = \frac{\sqrt{3}}{2}$ is 30 degrees i believe, im having toruble finding all the solutions to this problem
i also tried graphing ot sinf all of the solutions.

**mr fantastic** · June 30th 2008, 05:15 PM

Originally Posted by NoAsherelol

ih im sorry, this is my problem, $Sin\,x = \frac{\sqrt{3}}{2}$ is 30 degrees i believe, im having toruble finding all the solutions to this problem
i also tried graphing ot sinf all of the solutions.

We've gone from $\sin (3x) = \frac{\sqrt{3}}{2}$ to $\sin (2x) = \frac{\sqrt{3}}{2}$ to $\sin (x) = \frac{\sqrt{3}}{2}$ .

Which one is it?

**NoAsherelol** · June 30th 2008, 05:27 PM

Originally Posted by mr fantastic

We've gone from $\sin (3x) = \frac{\sqrt{3}}{2}$ to $\sin (2x) = \frac{\sqrt{3}}{2}$ to $\sin (x) = \frac{\sqrt{3}}{2}$ .

Which one is it?

its this $\sin (2x) = \frac{\sqrt{3}}{2}$ , listen man i am very stressed out, im sorry about the crazy post

**Jhevon** · June 30th 2008, 05:37 PM

Originally Posted by NoAsherelol

its this $\sin (2x) = \frac{\sqrt{3}}{2}$ , listen man i am very stressed out, im sorry about the crazy post

first thing you should know, is that if $\sin x = \frac {\sqrt{3}}2$ then $x = 60$ degrees or $x = 120$ degrees. that is only for $0 \le x \le 360$ . for all solutions you need something different. you need to tell us what solutions you are looking for. and furthermore, are you sure you shouldn't be working in radians, as opposed to degrees?

$\sin 2x = \frac {\sqrt{3}}2$

$\Rightarrow 2x = 60^o \mbox{ or } 120^o$ for $0^o \le x \le 360^o$

$\Rightarrow x = 30^o \mbox{ or } x = 60^o$ for $0^o \le x \le 360^o$

**ticbol** · June 30th 2008, 05:44 PM

Originally Posted by NoAsherelol

ih im sorry, this is my problem, $Sin\,x = \frac{\sqrt{3}}{2}$ is 30 degrees i believe, im having toruble finding all the solutions to this problem
i also tried graphing ot sinf all of the solutions.

mr fantastic got confused also, but he did not laugh. He has strong self-discipline.

Sorry if I laughed when you first wrote sin(3x) when you actually meant sin(2x).

So it is really for all the values of x when sin(x) = [sqrt(3)]/2.

sin(60degrees) = [sqrt(3)]/2. Not 30 degrees.

In one cycle, or in one revolution in the unit circle, or from 0 degrees up to 360 degees, there are only two quadrants where the sine value is positive. It is only in the 1st and 2nd quadrants.
In the 1st quadrant, the angle, or x here, is 60 degrees.
In the 2nd quadrant, x = 180 -60 = 120 degrees.

In the next revolution, or from 360deg up to 720 deg, there again only 2 possible x's.
In the 1st quadrant, x = 360 +60 = 420 deg
In the 2nd quadrant, x = 360 +120 = 480 deg.

In the next revolution, or from 720deg up to 1080deg,
1st uqdrant, x = 60 +720 = 780deg
2nd quadrant, x = 120 +720 = 840 deg.

Etc.

So for all the values of x in all revolutions,
x = 60 +n(360), and x = 120 +n(360), in degrees --------answer
where n = any non-negative integer.

**Jhevon** · June 30th 2008, 05:49 PM

memorize this table (i put the radian measures over the degree measures):

$\begin{array}{c|c|c|c} & \frac {\pi}6 & \frac {\pi}4 & \frac {\pi}3 \\ \theta & & & \\ & 30^o & 45^o & 60^o \\ \hline & & & \\ \sin \theta & \frac 12 & \frac {\sqrt {2}}2 & \frac {\sqrt {3}}2 \\ & & & \\ \hline & & & \\ \cos \theta & \frac {\sqrt{3}}2 & \frac {\sqrt{2}}2 & \frac 12 \\ & & & \end{array}$

that table tells you the sine and cosine of the special angles 30, 45 and 60 degrees (and note that, if you know the sine and cosine for an angle, you know the tangent of the angle as well. since tan(x) = sin(x)/cos(x)). all other special angles you can know from the graphs, which you should know anyway. your knowledge of reference angles allow you to deal with the analogous angles in the other quadrants (see ticbol's post)

here's the mnemonic for filling out the table. fill out the theta, sine and cosine columns. allow the first row, count up with the angles, 30, 45, 60. then, in the sine row, count up from 1 to 3, go the opposite way with cosine. then, divide everything by 2, then square root all the numerators, and you get that table

**NoAsherelol** · June 30th 2008, 06:20 PM

Originally Posted by Jhevon

first thing you should know, is that if $\sin x = \frac {\sqrt{3}}2$ then $x = 60$ degrees or $x = 120$ degrees. that is only for $0 \le x \le 360$ . for all solutions you need something different. you need to tell us what solutions you are looking for. and furthermore, are you sure you shouldn't be working in radians, as opposed to degrees?

$\sin 2x = \frac {\sqrt{3}}2$

$\Rightarrow 2x = 60^o \mbox{ or } 120^o$ for $0^o \le x \le 360^o$

$\Rightarrow x = 30^o \mbox{ or } x = 60^o$ for $0^o \le x \le 360^o$

as far as intervals go i dont know, it just asked me to find all solutions, and yes i think its in radians thanx

**Jhevon** · June 30th 2008, 06:28 PM

Originally Posted by NoAsherelol

as far as intervals go i dont know, it just asked me to find all solutions

that's fine. see ticbol's solution. he gives you all (that's what the n(360) is for).

and yes i think its in radians thanx

this, however, is something you should have said earlier, especially since you mentioned degrees and misled everyone. no worries though. just convert ticbol's solution to radians.

we go from degrees to radians by multiplying by $\frac {\pi}{180}$

**Reckoner** · June 30th 2008, 06:46 PM

Originally Posted by Jhevon

here's the mnemonic for filling out the table. fill out the theta, sine and cosine columns. allow the first row, count up with the angles, 30, 45, 60. then, in the sine row, count up from 1 to 3, go the opposite way with cosine. then, divide everything by 2, then square root all the numerators, and you get that table

I hadn't seen that mnemonic before. Thanks!

Originally Posted by NoAsherelol

as far as intervals go i dont know, it just asked me to find all solutions, and yes i think its in radians thanx

In this case we are looking for an angle (because the trigonometric functions are functions of angles), and since any given angle can be represented in infinitely many ways, the equation has infinitely many solutions.

One way to express the general solution set is to introduce an integer $n$ as a parameter:

We have $\sin2x = \frac{\sqrt3}2$ .

You should know (if not, listen to Jhevon!) that $\sin{\frac\pi3} = \sin60^\circ = \frac{\sqrt3}2$ . This means that 60° is one possible choice for the sought angle. So we have

$2x = 60^\circ\Rightarrow x = 30^\circ$

for quadrant I. If you look at the unit circle, you should see that there is another angle in quadrant II which, using a reference angle, you can determine as $180^\circ - 60^\circ = 120^\circ$ , so

$2x = 120^\circ\Rightarrow x = 60^\circ$

We have two choices for $x$ here. But we can add any multiple of 360° to an angle's measure without changing the angle it represents. So the general solution is

$2x = 60^\circ + 360^\circ n,\text{ or }2x = 120^\circ + 360^\circ n$

$\Rightarrow x = 30^\circ + 180^\circ n,\text{ or }x = 60^\circ + 180^\circ n,\;n\in\mathbb{Z}$

If you aren't familiar, the notation $n\in\mathbb{Z}$ just means that $n$ is an integer.

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