2sin2x = cos2x for - pi x pi
To solve the equation 2sin(2x) = cos(2x) for -π ≤ x ≤ π, we can use the double-angle formula sin(2x) = 2sin(x)cos(x) and rewrite the equation as:
2(2sin(x)cos(x)) = cos(2x)
Expanding the equation further:
4sin(x)cos(x) = cos(2x)
Using the double-angle formula again for cosine, cos(2x) = cos^2(x) - sin^2(x), we have:
4sin(x)cos(x) = cos^2(x) - sin^2(x)
Rearranging the equation:
4sin(x)cos(x) + sin^2(x) = cos^2(x)
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can substitute it into the equation:
4sin(x)cos(x) + sin^2(x) = 1 - sin^2(x)
Simplifying:
4sin(x)cos(x) + 2sin^2(x) - 1 = 0
Now we have a quadratic equation in terms of sin(x). Let's substitute y = sin(x):
4ycos(x) + 2y^2 - 1 = 0
Multiplying through by 2:
8y^2 + 4ycos(x) - 2 = 0
Using the quadratic formula, y = [-4cos(x) ± √(16cos^2(x) - 4(8)(-2))] / (2(8))
Simplifying:
y = [-4cos(x) ± √(64cos^2(x) + 64)] / 16
y = [-cos(x) ± √(cos^2(x) + 1)] / 2
Now substitute y back as sin(x):
sin(x) = [-cos(x) ± √(cos^2(x) + 1)] / 2
We can solve for sin(x) by considering the two cases of + and -:
1) sin(x) = [-cos(x) + √(cos^2(x) + 1)] / 2
Rearranging:
cos(x) = -[1 + 2sin(x)] / √(4sin^2(x) + 1)
Since -π ≤ x ≤ π, we can use the unit circle to determine where the cosine values are negative or positive. From the unit circle, we can see that the cosine in the given range is positive at x = -π/2 and x = π/2.
For x = -π/2:
cos(-π/2) = -[1 + 2sin(-π/2)] / √(4sin^2(-π/2) + 1)
0 = -1 / √(4 + 1)
This is not true, so x = -π/2 is not a valid solution.
For x = π/2:
cos(π/2) = -[1 + 2sin(π/2)] / √(4sin^2(π/2) + 1)
0 = -1 / √(4 + 1)
This is not true, so x = π/2 is not a valid solution.
Therefore, the equation 2sin(2x) = cos(2x) has no solutions in the range -π ≤ x ≤ π.
To solve the equation 2sin(2x) = cos(2x) for -π ≤ x ≤ π, we can use trigonometric identities.
Step 1: Let's write the equation using the double-angle identity for sine and cosine.
2(2sin(x)cos(x)) = cos^2(x) - sin^2(x)
Step 2: Simplify by expanding the right side of the equation using the Pythagorean identity for sine and cosine.
4sin(x)cos(x) = cos^2(x) - (1 - cos^2(x))
4sin(x)cos(x) = 2cos^2(x) - 1
Step 3: Rearrange the equation by bringing all terms to one side to set it equal to zero.
4sin(x)cos(x) - 2cos^2(x) + 1 = 0
Step 4: Factor out a common factor of 2.
2(2sin(x)cos(x) - cos^2(x) + 1/2) = 0
Step 5: Notice that the expression inside the parentheses can be rewritten as (cos(x) - 1/2)^2.
2(2sin(x)cos(x) - (cos(x) - 1/2)^2) = 0
Step 6: Apply the zero product property, meaning that if the product of factors is zero, then at least one of the factors must be zero.
2sin(x)cos(x) - (cos(x) - 1/2)^2 = 0
Now we have two cases to consider:
Case 1: Setting the first factor equal to zero:
2sin(x)cos(x) = 0
This equation is true when either sin(x) = 0 or cos(x) = 0.
For sin(x) = 0:
x = 0, π
For cos(x) = 0:
x = π/2, -π/2
Case 2: Setting the second factor equal to zero:
(cos(x) - 1/2)^2 = 0
Solving for cos(x) - 1/2 = 0 gives:
cos(x) = 1/2
This equation is true when:
x = π/3, -π/3
Putting all the solutions together, we have:
x = 0, π/2, -π/2, π/3, -π/3
To solve the equation 2sin(2x) = cos(2x) for -π ≤ x ≤ π, we can use the trigonometric identity sin^2(x) + cos^2(x) = 1.
Step 1: Rewrite the equation using the identity:
2sin(2x) = cos(2x)
2(2sin(x)cos(x)) = cos(2x)
4sin(x)cos(x) = cos(2x)
Step 2: Use the double angle formula for cosine cos(2x) = 2cos^2(x) - 1:
4sin(x)cos(x) = 2cos^2(x) - 1
Step 3: Rearrange the equation:
2cos^2(x) - 4sin(x)cos(x) - 1 = 0
Step 4: Factoring the quadratic equation:
(2cos(x) + 1)(cos(x) - 1) = 0
Step 5: Set each factor equal to zero and solve for x:
2cos(x) + 1 = 0
cos(x) = -1/2
x = π/3, 5π/3
cos(x) - 1 = 0
cos(x) = 1
x = 0
So the solutions for -π ≤ x ≤ π are: x = 0, π/3, 5π/3.