identify all solutions in the interval [0,2pi]: cos (2x) csc^2x=2cos(2x)
I have no clue how to solve.
thanks
To solve the equation cos(2x) csc^2x = 2cos(2x) in the interval [0, 2π], we will follow these steps:
Step 1: Simplify the equation.
Using the trigonometric identity csc^2x = 1/sin^2x, we can rewrite the equation as cos(2x) / sin^2x = 2cos(2x).
Step 2: Multiply both sides of the equation by sin^2x to eliminate the denominators:
cos(2x) = 2cos(2x) * sin^2x.
Step 3: Simplify further:
cos(2x) = 2cos(2x) * (1 - cos^2x).
Step 4: Expand the equation:
cos(2x) = 2cos(2x) - 2cos^3(2x).
Step 5: Move all terms to one side of the equation:
cos(2x) - 2cos(2x) + 2cos^3(2x) = 0.
Step 6: Combine like terms:
-cos(2x) + 2cos^3(2x) = 0.
Step 7: Factor out cos(2x):
cos(2x) (-1 + 2cos^2(2x)) = 0.
Step 8: Set each factor equal to 0 and solve for x:
cos(2x) = 0. This occurs when cos(2x) = 0, so x = π/4 and x = 3π/4.
-1 + 2cos^2(2x) = 0. To solve this, rearrange the equation as cos^2(2x) = 1/2 and take the square root of both sides: cos(2x) = ±√(1/2). This occurs when 2x = π/4, 7π/4, 3π/4, 5π/4. Therefore, x = π/8, 7π/8, 3π/8, 5π/8.
So, the solutions in the interval [0, 2π] are x = π/4, 3π/4, π/8, 5π/8, 3π/8, 7π/8.
To solve the equation cos(2x) csc^2(x) = 2cos(2x) in the interval [0, 2π], we can follow these steps:
Step 1: Simplify the equation.
Using trigonometric identities, we can simplify the equation:
cos(2x) csc^2(x) = 2cos(2x)
cos(2x) (1/sin^2(x)) = 2cos(2x)
cos(2x)/sin^2(x) = 2cos(2x)
Step 2: Eliminate the denominator.
Since sin^2(x) cannot be zero (as it is in the denominator), we can multiply sin^2(x) to both sides to get rid of the fraction:
cos(2x) = 2cos(2x) * sin^2(x)
cos(2x) = 2sin^2(x)cos(2x)
Step 3: Solve for cos(2x) = 0.
cos(2x) = 0 is one of the possible solutions. To find the values of x that satisfy this condition, we can solve:
2x = π/2 + nπ, where n is an integer
x = (π/2 + nπ)/2, where n is an integer
Step 4: Solve for cos(2x) ≠ 0.
For cos(2x) ≠ 0, we can divide both sides of the equation by cos(2x):
1 = 2sin^2(x)
sin^2(x) = 1/2
sin(x) = ± √(1/2)
sin(x) = ± 1/√2
sin(x) = ± √2/2
Step 5: Find the values of x for sin(x) = ± √2/2.
We know that sin(x) = ± √2/2 for x = π/4 and x = 3π/4 (among other values), because sin(π/4) = √2/2 and sin(3π/4) = √2/2.
Now, we have the following potential solutions:
x = π/4 + nπ, where n is an integer
x = 3π/4 + nπ, where n is an integer
x = (π/2 + nπ)/2, where n is an integer
Combining all the potential solutions, the solutions in the interval [0, 2π] for the equation cos(2x) csc^2(x) = 2cos(2x) are:
- x = π/4, 3π/4, π/2, 3π/2, π, 2π (and their respective values for n = ±1, ±2, ±3, ...)
These values satisfy the original equation in the given interval.
cos(2x) csc^2(x) = 2cos(2x)
well, duh - divide by cos(2x), assuming cos(2x)≠0
csc^2(x) = 2
csc(x) = ±√2
x = all odd multiples of π/4
But, at those values of x, cos(2x)=0. So that set of roots is verboten.
Naturally the equation is true if cos(2x) = 0, meaning 2x is an odd multiple of π/2. Or, again, x is an odd multiple of π/4.
http://www.wolframalpha.com/input/?i=cos%282x%29+csc%5E2%28x%29+%3D+2cos%282x%29