Find general solution for the following

1.

2.

can anyone help?

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- September 16th 2007, 11:45 PMacc100jtQuestions on trigo. equation
Find general solution for the following

1.

2.

can anyone help? - September 17th 2007, 04:35 AMJhevon
- September 17th 2007, 10:57 AMred_dog
- September 17th 2007, 11:56 AMticbol
Umm, these are really very hard to solve. Need tricks.

Even when I was yet in High School, I resort to iteration, or trial and error, in solving equations that I didn't know how to solve the usual ways.

Problem 1 is is a good candidate for my iteration.

By luck, or because of the sqrt(3) in the equation, I tried first, x = 30degrees.

cos(3*30deg) =? 2cos(30deg) -sqrt(3)*sin(3*30deg)

cos(90deg) =? 2cos(30deg) -sqrt(3)*sin(90deg)

0 =? 2[sqrt(3) / 2] -sqrt(3)*(1)

0 =? sqrt(3) -sqrt(3)

0 =? 0

Yes, so x = 30deg is OK.

I was just lucky to hit it in one try only. So, 30deg is a solution. There might be some more but I don't know how, or I don't want, to continue after that.

------------------------------

For the second problem, another trick.

cos(2x +60deg) = sin(x) ----------(ii)

Expanding that will lead to an equation that is as complicated as the first one, so another dip into our bag of tricks:

We know that cosA = sin(90deg -A).

That means A +(90deg -A) = 90 deg.

Or, cosA = sinB -----if angles A and B are complementary.

So, using that on (ii),

(2x +60deg) +x = 90deg

3x = 90 -60

x = 30/3

x = 10 deg ------------a solution.

Check,

cos(2*10deg +60deg) =? sin(10deg)

cos(80deg) =? sin(10deg)

Using a calculator,

0.173648178 =? 0.173648178

Yes, so, x = 10 degrees is correct. - September 17th 2007, 08:01 PMJhevon