solve for all real#
sin2x(csc2x-2)=0
To solve the equation sin(2x)(csc(2x) - 2) = 0 for all real numbers x, we need to find the values of x where the equation holds true.
First, let's simplify the equation by factoring out sin(2x):
sin(2x)(csc(2x) - 2) = 0
We know that sin(2x) = 0 for all values of x where 2x is an integer multiple of π. So, one solution is:
2x = nπ, where n is an integer.
Now, let's consider the second part of the equation: csc(2x) - 2 = 0
We can rewrite this as:
csc(2x) = 2
Since csc(2x) is the reciprocal of sin(2x), we have:
1/sin(2x) = 2
Take the reciprocal of both sides:
sin(2x) = 1/2
Now, we look for the values of x where sin(2x) = 1/2. This occurs when 2x is equal to π/6, 5π/6, 13π/6, etc. We can represent this as:
2x = π/6 + nπ, where n is an integer.
Divide both sides by 2:
x = π/12 + nπ/2, where n is an integer.
Combining both solutions, we have two possible sets of values for x:
1) 2x = nπ, where n is an integer.
Divide both sides by 2:
x = nπ/2, where n is an integer.
2) x = π/12 + nπ/2, where n is an integer.
So, the equation sin(2x)(csc(2x) - 2) = 0 holds true for any real number x that satisfies either x = nπ/2 or x = π/12 + nπ/2, where n is an integer.