Thread: Questions on trigo. equation

Originally Posted by acc100jt

Find general solution for the following
1.
2.

can anyone help?

expand using the addition and double angle formulas and then see what you can do.

recall:








note: and

Originally Posted by acc100jt

Find general solution for the following
1.

Divide both members by 2:






Now, continue.

Originally Posted by acc100jt

Find general solution for the following
1.
2.

can anyone help?

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.

Originally Posted by red_dog

Divide both members by 2:






Now, continue.

nice!