trigo solving

**thereddevils** · October 2nd 2009, 04:09 AM

Find the value of x of the following trigonometric equation with x is between -pi and pi

| sin 2x | = 1/2

My work

sin 2x = 1/2 (case 1)

2x= 30 , 150

x= 15 and 75

sin (-2x) = 1/2

There is no solution for this .

case 2

sin 2x = -1/2

no solution

sin (-2x) = -1/2

-2x = 30 , 150

x = -15 , -75

Am i correct ??

**red_dog** · October 2nd 2009, 07:14 AM

You have just two cases:

$\sin 2x=\frac{1}{2}$ and $\sin 2x=-\frac{1}{2}$

**thereddevils** · October 2nd 2009, 07:24 AM

Originally Posted by red_dog

You have just two cases:

$\sin 2x=\frac{1}{2}$ and $\sin 2x=-\frac{1}{2}$

so the solution will be just pi/4 and 5pi/12 but this is for the range 0 and pi only . how about -pi and 0 ?

THanks .

**Grandad** · October 2nd 2009, 08:04 AM

Hello thereddevils

As red_dog has said, you need to solve two equations

$\sin2x = \frac12$

and

$\sin2x=-\frac12$

Since you want values of $x$ between $-\pi$ and $\pi$ , you'll need to look at values of $2x$ between $-2\pi$ and $2\pi$ .

I always find the graph of $\sin\theta$ easier to use than the unit circle, so look at the sketch I've attached. The easiest value is, of course, $\sin(\tfrac{\pi}{6})=\tfrac12$ . Look at where the dotted lines intersect the graph. They are at:

$-\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{7 \pi}{6},\frac{11\pi}{6}$

So these are the possible values of $2x$ . Divide them all by $2$ , and you're done.

Grandad

**thereddevils** · October 2nd 2009, 08:30 AM

Originally Posted by Grandad

Hello thereddevils

As red_dog has said, you need to solve two equations

$\sin2x = \frac12$

and

$\sin2x=-\frac12$

Since you want values of $x$ between $-\pi$ and $\pi$ , you'll need to look at values of $2x$ between $-2\pi$ and $2\pi$ .

I always find the graph of $\sin\theta$ easier to use than the unit circle, so look at the sketch I've attached. The easiest value is, of course, $\sin(\tfrac{\pi}{6})=\tfrac12$ . Look at where the dotted lines intersect the graph. They are at:

$-\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{7 \pi}{6},\frac{11\pi}{6}$

So these are the possible values of $2x$ . Divide them all by $2$ , and you're done.

Grandad

Thanks Grandad ,

sin 2x =1/2

so 2x = $\frac{\pi}{6},\frac{5\pi}{6}$

sin 2x = -1/2

$2x= \frac{7\pi}{6},\frac{11\pi}{6}$

How to get the negative one ??

Thanks .

**Grandad** · October 2nd 2009, 12:26 PM

Originally Posted by thereddevils

How to get the negative one ??

Thanks .

Use the rotational symmetry of the graph. The negative part is just a $180^o$ rotation of the positive part. So where you have $\pi/6$ on the right-hand side, you'll have $-\pi/6$ on the left.

Grandad

**thereddevils** · October 2nd 2009, 07:00 PM

Originally Posted by Grandad

Use the rotational symmetry of the graph. The negative part is just a $180^o$ rotation of the positive part. So where you have $\pi/6$ on the right-hand side, you'll have $-\pi/6$ on the left.

Grandad

THanks Grandad , does this also work for cos and tan ?

**Grandad** · October 2nd 2009, 09:55 PM

Hello thereddevils

Originally Posted by thereddevils

THanks Grandad , does this also work for cos and tan ?

Absolutely! Just sketch the graph, then use the section between $0$ and $\pi/2$ (the bit I've indicated below inside the red rectangle) as a 'building block', rotating and reflecting it as necessary, to create the whole graph, for any range of values you want. Then, provided you know the angle in the range $0$ to $\pi/2$ , you can easily work out the corresponding angle in any other range.

Grandad

**thereddevils** · October 3rd 2009, 02:37 AM

Originally Posted by Grandad

Hello thereddevilsAbsolutely! Just sketch the graph, then use the section between $0$ and $\pi/2$ (the bit I've indicated below inside the red rectangle) as a 'building block', rotating and reflecting it as necessary, to create the whole graph, for any range of values you want. Then, provided you know the angle in the range $0$ to $\pi/2$ , you can easily work out the corresponding angle in any other range.

Grandad

THanks Grandad , i see . Can i bring in another question which i asked in another thread for reference .
Solve for x for the range of x between -pi and pi
sin x = cos x

tan x=1

x=45

Would the other x be -45 ?? I don think so ..

**Grandad** · October 3rd 2009, 07:27 AM

Hello thereddevils

Originally Posted by thereddevils

THanks Grandad , i see . Can i bring in another question which i asked in another thread for reference .
Solve for x for the range of x between -pi and pi
sin x = cos x

tan x=1

x=45

Would the other x be -45 ?? I don think so ..

Have a look at the diagram I've attached, which shows a sketch-graph of $y = \tan x$ , between $\pm360^o$ , and the line $y = 1$ as a dotted line.

You'll see that I've drawn a rectangle around the section of the graph between $x = 0$ and $x = 90^o$ - rectangle number (1). This rectangle is then rotated about the point $(90,0)$ to form rectangle (2). And it has been translated through $180^o$ to form rectangle (3) ... And so on, to create as many sections of the graph as you need.

Now we know that the first value where $\tan x = 1$ is $x = 45^o$ . Using this graph you can see that there are others where $x = 180 +45$ and $-180+45$ , $-360 + 45$ , ... So the solutions between $\pm180^o$ are $-135^o, 45^o$ .

We can also work out in the same way where $\tan x = -1$ . Rectangle (2) shows that there will be one value at $x = 180-45 = 135^o$ . And there'll be others at $x = 360-45 = 315^o,-45^o, -180-45 = -225^o, ...$ etc.

Provided you can sketch the graphs of sine, cosine and tangent quickly - and it's worth learning them - I think you'll find this is a great help in solving trig equations.

Grandad

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