Solve:
cos(2x-180) - sin(x-90)=0
my work:
cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx
-cos2x - sin2x= cosx
-cos^2x + sin^2x -2sinxcosx=cosx
I'm stuck here. I tried subtracting cosx from both sides and making sin^2x into 1- cos^2x, but I still can't seem to solve the problem.
To solve the equation cos(2x-180) - sin(x-90) = 0, let's break it down step by step.
1. Rewrite the equation using double angle identities:
cos(2x)cos(180) + sin(2x)sin(180) = sin(x)cos(90) - sin(90)cos(x)
2. Simplify the trigonometric values:
-1 * cos(2x) - 0 * sin(2x) = 0 - 1 * cos(x)
3. Simplify the equation further:
-cos(2x) = -cos(x)
4. Bring both the cosine terms to the left-hand side of the equation:
cos(2x) - cos(x) = 0
5. Apply the cosine identity (cos(a) - cos(b) = -2sin((a+b)/2) * sin((a-b)/2)):
-2sin((2x + x)/2) * sin((2x - x)/2) = 0
6. Simplify the equation:
-2sin(3x/2) * sin(x/2) = 0
Now, we have two possibilities:
1. -2sin(3x/2) = 0
To solve this equation, set sin(3x/2) = 0:
sin(3x/2) = 0
We know that sin(θ) = 0 when θ = nπ (where n is an integer). Therefore, we have:
3x/2 = nπ
3x = 2nπ
x = (2/3)nπ
2. sin(x/2) = 0
To solve this equation, set x/2 = nπ:
x = 2nπ
Therefore, the possible solutions are:
x = (2/3)nπ and x = 2nπ, where n is an integer.