Trigonometry?

Question

Find the general solutions:
3*sin(3x)-1=cos(6x)

Answer 1

Double angle formula for cosine gives
cos(6x)=cos(3x)^2-sin(3x)^2
and also
cos(3x)^2=1-sin(3x)^2.
So our equation simplifies to
3sin(3x)-1=1-2sin(3x)^2
or
2sin(3x)^2+3sin(3x)-2=0
Factoring,
(2sin(3x)-1)(sin(3x)+2)=0
Since sin(3x) cannot equal -2, we must have sin(3x)=1/2.
This occurs when 3x=(-1)^k*pi/6+k*pi where k is any integer. Thus, x=(-1)^k*pi/18+k*pi/3 where k is any integer.

Answer 2

I would let 3x = Z and 6x = 2Z. This way, you can rewrite the equation to be 3sin(z)-1=cos(2z), then use double number identity.

Once the equation is solved, translate it back the answre in term of x.

Answer 3

3 sin(3x) - 1 = cos(6x)

3 sin(3x) - 1 = cos(2*3x)

3sin(3x) -1 = 1 - 2sin^2(3x)

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

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

2sin^2(3x) + 4 sin(3x) - sin(3x) - 2 = 0

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

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

since sin value is between -1 and + 1,

the only possibilty solution is

2sin(3x) - 1 = 0

2 sin(3x) = 1

sin(3x) = 1/2

3x = pi/6

x = pi/18 or 17pi/18