Trig help

[1141]

Can anybody help?

I've got these questions that I need to do, but I don't understand how to do them and my text book is really bad at explaining things:

Find the roots in the interval -180 =< x =< -180 of each of the following equations:

a.) cos2x = 0.246
b.) sin3x = -0.2


can anybody explain to me how I'm supposed to do this??

 

[galactus]

Can anybody help?

I've got these questions that I need to do, but I don't understand how to do them and my text book is really bad at explaining things:

Find the roots in the interval \(\displaystyle -180\leq x \leq180\) of each of the following equations:

[quote:jzku89na]a.) cos2x = 0.246

[/quote:jzku89na]

Solve for x, \(\displaystyle x=\frac{45\left(4C{\pi}\pm\frac{cos^{-1}(\frac{123}{500}){\pi}}{90}\right)}{\pi}\)

Let C=0, 1, ..... and see what roots are in the required interval.

The above is the general form. We can solve a for x and get \(\displaystyle x=\pm\frac{cos^{-1}(\frac{123}{500})}{2}\) and use the periodicity of cosine to find the other angles.

x becomes -37.88 and 37.88 if we add 180 to -37.88, we get 142.12

If we add 180 to 37.88, we are outside the interval.

 

[soroban]

Hello, 1141!

Another approach . . .

Find the roots in the interval \(\displaystyle \text{-}180^o \leq x \leq 180^o\)

. . \(\displaystyle a)\;\;\cos2x \:=\: 0.246\)

\(\displaystyle \text{We have: }\:2x \;=\;\cos^{-1}(0.246)\)

. . . . . . . .\(\displaystyle x \;=\;\tfrac{1}{2}\cos^{-1}(0.246)\)

. . . . . . . .\(\displaystyle x \;\approx\;\pm37.88^o\)

. . \(\displaystyle x\text{ is in all four quadrants.}\)


\(\displaystyle \text{Therefore: }\:x \;=\;\pm37.88^o,\:\pm142.12^o\)

 

[1141]

Hello, 1141!

Another approach . . .

Find the roots in the interval \(\displaystyle \text{-}180^o \leq x \leq 180^o\)

. . \(\displaystyle a)\;\;\cos2x \:=\: 0.246\)

\(\displaystyle \text{We have: }\:2x \;=\;\cos^{-1}(0.246)\)

. . . . . . . .\(\displaystyle x \;=\;\tfrac{1}{2}\cos^{-1}(0.246)\)

. . . . . . . .\(\displaystyle x \;\approx\;\pm37.88^o\)

. . \(\displaystyle x\text{ is in all four quadrants.}\)

\(\displaystyle \text{Therefore: }\:x \;=\;\pm37.88^o,\:\pm142.12^o\)

okay, I understand that. I tried applying that method to the other question, of sin3x = -0.12 but I keep getting the wrong answer. I keep getting -2.29. But the answers in my text book are : -176.2, -123.8, -56.2, -3.8, 63.8, 116.2.

I don't understand how those answers were found??

 

[galactus]

Make sure your calculator is not in Radian mode. If it is, put it in Degrees.

 

[1141]

Make sure your calculator is not in Radian mode. If it is, put it in Degrees.

That's the first thing I checked. It's in Degrees.

 

[galactus]

As a start, solve the given for x.

\(\displaystyle x=\frac{sin^{-1}(-.2)}{3}=-3.845653....\)

Go from there using the same idea Soroban outlined.

 

[1141]

As a start, solve the given for x.

\(\displaystyle x=\frac{sin^{-1}(-.2)}{3}=-3.845653....\)

Go from there using the same idea Soroban outlined.

Alright, I worked it out to find the root -3.8 and then -176.2, but how would I find the other roots of -123.8, -56.2, 63.8 and 116.2??

 

[mmm4444bot]

?
Use the periodicity of the sine function, to find repeating solutions.

Here's a quick example. (I'll work with degrees, too, instead of the usual radians.)

Solve:

2 sin(3x) - 1 = 0

First, we solve for 3x in one period [0, 360°).

sin(3x) = 1/2

3x = 30°

or

3x = 150°

To get all solutions, we add multiples of 360°, yes?

3x = 30° + 360° k

or

3x = 150° + 360° k

where k is an Integer.

Now, we solve these two equations for x.

x = 10° + 120° k

or

x = 50° + 120° k

where k is an Integer

You can follow the same strategy, followed by picking values of k until x falls outside each boundary of the interval in this exercise.

 

[1141]

Re:

?
Use the periodicity of the sine function, to find repeating solutions.

Here's a quick example. (I'll work with degrees, too, instead of the usual radians.)

Solve:

2 sin(3x) - 1 = 0

First, we solve for 3x in one period [0, 360°).

sin(3x) = 1/2

3x = 30°

or

3x = 150°

To get all solutions, we add multiples of 360°, yes?

3x = 30° + 360° k

or

3x = 150° + 360° k

where k is an Integer.

Now, we solve these two equations for x.

x = 10° + 120° k

or

x = 50° + 120° k

where k is an Integer

You can follow the same strategy, followed by picking values of k until x falls outside each boundary of the interval in this exercise.

Thank you. I understand that now.
Do you think you could help with these questions:

a.) 10sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? - 5cos[sup:3vq2tb5a]2[/sup:3vq2tb5a]? + 2 = 4sin?

b.) 4sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? cos? = tan[sup:3vq2tb5a]2[/sup:3vq2tb5a]?

For a.) this is my working out so far:

10sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? - 5cos[sup:3vq2tb5a]2[/sup:3vq2tb5a]? + 2 = 4sin?
10sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? - 5(1-sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? ) + 2 = 4sin?
10sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? - 5 + 5sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? + 2 = 4sin?
15sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? - 3 = 4sin?
15sin[sup:3vq2tb5a]2[/sup:3vq2tb5a]? - 4sin? - 3 = 0

I don't think that I can factorize it, and I tried finding the roots like you would normally do with a quadratic equation they came out as 0.6 and -0.3333... . I'm not sure what I have to do next.

As for b.) I have no idea how to even start it.

 

[Deleted member 4993]

Re: Re:

Thank you. I understand that now.
Do you think you could help with these questions:

Start a new thread - with a new problem

a.) 10sin[sup:1euf1vz1]2[/sup:1euf1vz1]? - 5cos[sup:1euf1vz1]2[/sup:1euf1vz1]? + 2 = 4sin?

b.) 4sin[sup:1euf1vz1]2[/sup:1euf1vz1]? cos? = tan[sup:1euf1vz1]2[/sup:1euf1vz1]?

For a.) this is my working out so far:

10sin[sup:1euf1vz1]2[/sup:1euf1vz1]? - 5cos[sup:1euf1vz1]2[/sup:1euf1vz1]? + 2 = 4sin?
10sin[sup:1euf1vz1]2[/sup:1euf1vz1]? - 5(1-sin[sup:1euf1vz1]2[/sup:1euf1vz1]? ) + 2 = 4sin?
10sin[sup:1euf1vz1]2[/sup:1euf1vz1]? - 5 + 5sin[sup:1euf1vz1]2[/sup:1euf1vz1]? + 2 = 4sin?
15sin[sup:1euf1vz1]2[/sup:1euf1vz1]? - 3 = 4sin?
15sin[sup:1euf1vz1]2[/sup:1euf1vz1]? - 4sin? - 3 = 0

I don't think that I can factorize it, and I tried finding the roots like you would normally do with a quadratic equation they came out as 0.6 and -0.3333... . I'm not sure what I have to do next.

You are correct upto here. Now get your calculator out and use sin[sup:1euf1vz1]-1[/sup:1euf1vz1] function button to do the rest.

As for b.) I have no idea how to even start it.

[/quote]

b.) 4sin[sup:1euf1vz1]2[/sup:1euf1vz1]? cos? = tan[sup:1euf1vz1]2[/sup:1euf1vz1]? [/i]

\(\displaystyle 4*sin^2(\theta) * cos(\theta) = \frac{sin^2(\theta)}{cos^2(\theta)}\)

\(\displaystyle 4*sin^2(\theta) * cos^3(\theta) = sin^2(\theta)\)

\(\displaystyle 4*sin^2(\theta) * cos^3(\theta) - sin^2(\theta) = 0\)

Now continue...