Question

Cos9*x-cos5*x=√3sin2*x

Answer 1

-2sin(9x+5x/2)*sin(9x-5x/2)=√3sin2x
-2sin7x*sin2x=√3sin2x
-2sin7x*sin2x-√3sin2x=0
sin2x(-2sin7x-√3)=0
1) sin2x=0
2x=pi*n
x=pi*n/2
2) -2sin7x-√3=0
-2sin7x=√3
-sin7x=√3/2
sin7x=-√3/2
7x=(-1)^n+1 * pi/3+pi*n
x=(-1)^n+1 * pi/21+pi*n/7

Answer 2

This is a trigonometric equation that can be solved by using some identities and techniques. Here are the steps to find the solution:

1. Rewrite the equation as cos(9x) - cos(5x) - sqrt(3)sin(2x) = 0. 2. Use the identity cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2) to rewrite the left-hand side as -2sin(7x)sin(2x) - sqrt(3)sin(2x) = 0. 3. Factor out sin(2x) and get sin(2x)(-2sin(7x) - sqrt(3)) = 0. 4. Use the zero product property and set each factor equal to zero. This gives sin(2x) = 0 and -2sin(7x) - sqrt(3) = 0. 5. Solve for x in each equation. For sin(2x) = 0, we get x = k*pi/2, where k is any integer. For -2sin(7x) - sqrt(3) = 0, we get x = (1/7)arcsin(sqrt(3)/2) + (2k+1)pi/14, where k is any integer. 6. Combine the solutions and write them in the form x = a + b*k, where a and b are constants and k is any integer. The final answer is x = pi/2 + k*pi or x = (1/7)arcsin(sqrt(3)/2) + (2k+1)pi/14.