hard math problem?

Question

find to the nearest degree all values of the interval 0 degrees 3cos2@+sin@ - 1=0.

I think it is one of the cos 2 A identities, but theres three, and Im not sure

Answer 1

I'll use x instead of theta
3cos 2x + sin x - 1 = 0
3(1 - 2sin^2 x) + sin x -1 = 0
3 - 6sin^2 x + sin x -1 = 0
-6sin^2 x + sin x +2 = 0
6sin^2 x - sin x - 2 = 0
(2sin x + 1)(3sin x - 2) = 0
sin x = -1/2 sin x = 2/3
x = 210, 330 x = 41.81, 138.19
x = 41.81, 138.19, 210, 330

Answer 2

Yep, you need the cosine double angle formula to solve this one. cos2@ = (cos@)^2 - (sin@)^2.
Thus you now have
3[(cos@)^2 - (sin@)^2]+sin@-1=0
From the Pythagorean Theorem, you have (cos@)^2+(sin@)^2 = 1 so you can replace
(cos@)^2 with 1-(sin@)^2
Distribute and collect terms and you have
3-6(sin@)^2+sin@-1=0
Thus
-6(sin@)^2+sin@+2=0
This is a quadratic in sin@, so use the quadratic formula.
sin@ = [-1+/- sqrt(1^2-4(-6)(2)]/[(2*-6)]
sin@ = [-1+/- sqrt(49)]/-12
sin@ = [-1+/-7]/-12
sin@ = 6/-12 or sin@ = -8/-12
sin@ = -1/2 or sin@ = 2/3
Pop these numbers in your calculator using arcsines and don't forget to find all angles between 0 and 360 that satisfy the equation.
Have fun!
Hint: sin@=-1/2 means @=(5/6)*pi or (7/6)*pi which is all there are in the 2nd and third quadrants.
Do the same for sin@=2/3

Answer 3

3cos(2x) + sin(x) - 1 = 0

I'd use cos(2x) = 1 - 2sin^2(x). That makes the original equation become

3(1 - 2sin^2(x)) + sin(x) - 1 = 0
3 - 6(sin(x))^2 + sin(x) - 1 = 0
6(sin(x))^2 - sin(x) - 2 = 0

Then it becomes a polynomial algebra problem. You can factor this expression, treating sin(x) as a variable.

(3sin(x) - 2)(2sin(x) + 1) = 0

3sin(x) - 2 = 0 or 2sin(x) + 1 = 0
sin(x) = 2/3 or sin(x) = -1/2

Each of these equations has two solutions for x in the given domain.

Answer 4

cos(2a) = 1 - 2sin^2(a)

So

3 - 6sin^2(a) + sin(a) - 1 = 0
-6sin^2(a) + sin(a) + 2 = 0
6sin^2(a) - sin(a) - 2 = 0

Let's rewrite this with x = sin(a)

6x^2 - x - 2 = 0

x = (1 +/- sqrt(1 + 4 * 6 * 2))/(2 * 6)
x = (1 +/- sqrt(1 + 48))/(12)
x = (1 +/- sqrt(49))/12
x = (1 +/- 7)/12
x = 8/12 or x = -6/12
x = 2/3 or x = -1/2

so

sin(a) = 2/3 or sin(a) = -1/2
a = 41.8 or a = -30 or 330

Also, a = 180 - 41.8 or 138.2 and a = 180 + 30 = 210