Solving trigonometric equations.

Hello

The question was:

Solve the following on the interval . Give your answers to 3 sf.

I was able to solve this using the harmonic form, expressing the in terms of . I would like to know if it is possible, better, to solve this using identities, needless to say I can't see how and would appreciate a hint as how to start.

Thank you.

Re: Solving trigonometric equations.

Quote:

Originally Posted by

**Furyan** Hello

The question was:

Solve the following on the interval

. Give your answers to 3 sf.

I was able to solve this using the harmonic form, expressing the

in terms of

. I would like to know if it is possible, better, to solve this using identities, needless to say I can't see how and would appreciate a hint as how to start.

Thank you.

can you finish? don't forget to check for extraneous roots.

Re: Solving trigonometric equations.

Hi Furyan! :)

You would need the right trigonometric identity:

List of trigonometric identities - Linear combinations - Wikipedia

Edit: Oh never mind, I'm guessing that this is what you meant with the harmonic form.

Re: Solving trigonometric equations.

Re: Solving trigonometric equations.

Hello ILikeSerena

Quote:

Originally Posted by

**ILikeSerena**

Thank you. Yes I think that's what my book calls the 'harmonic form'. I was hoping there was an easier way, but maybe not. Thanks for your efforts all the same. :)

Re: Solving trigonometric equations.

To answer your question about extraneous roots, substitution into the original equation will in general show if the solution is extraneous or not.

I would have solved the equation much the same way you did:

First solution:

Observing that in a right triangle, the two inverse trig. functions represent complementary angles, we have:

Adding to get the angle in the desired range, we get:

Second solution:

We could plug this into a calculator, but there is a slightly simpler representation.

Construct a right triangle with horizontal base 1, altitude 2 and hypotenuse with the right angle at the lower right, then from the lower left vertex draw a line segment to the vertical leg at the point 1/2 unit above the right angle, and we have the angle subtended by this segment and the hypotenuse as . Using the law of cosines, we find:

Re: Solving trigonometric equations.

2 Attachment(s)

Re: Solving trigonometric equations.

Attachment 23169

Notice we have:

thus:

Attachment 23170

Notice that

Using the law of cosines allows you to find .

Re: Solving trigonometric equations.

Hello MarkFL2(Bow)

Quote:

Originally Posted by

**MarkFL2**

It's extremely kind of you to go to the trouble to post that. It's very helpful, now I see where you were getting , nice triangles. To be honest, at the moment I don't understand everything you've done, but I see that if I can it's going to make this topic very much more engaging. I'm going to look at it in more detail and get back to you if I have any questions. If that's alright.

Thanks again for all your help :)

Re: Solving trigonometric equations.

I just noticed a much simpler method to find . Rather than using the law of cosines, we may use the law of sines as well:

Multiplying through by 3/2 and recalling , we have:

All of this really points out that the method used by **skeeter** is the way to go. (Cool)

Re: Solving trigonometric equations.

Hi MarkFL2

Quote:

Originally Posted by

**MarkFL2** I just noticed a much simpler method to find

. Rather than using the law of cosines, we may use the law of sines as well:

Multiplying through by 3/2 and recalling

, we have:

All of this really points out that the method used by

**skeeter** is the way to go. (Cool)

Thank you for that. I think you are right, the method used by skeeter is definately the way to go in this case and I used it to successfully solve other similiar question. However there are questions in which it states you have to solve equations of the form using the method which the book calls the 'harmonic form', of which your method seems to be an extension. Both make use of all three trigonometric ratios in a right angle triangle. Although I can use the method in the book I don't understand how it works. I think if I look at it more closely together with what you have shown me it's going to make a lot more sense and perhaps I can use it to find exact values, where they exist, as you did.

Thanks again for your efforts. :)