sin2xcosx-sinx=0 interval(0,2pi)
Use double angle formula to reduce sin(2x) to 2sin(x)cos(x):
2sin(x)cos(x)cos(x)-sin(x)=0
sin(x)(2cos²(x)-1)=0
=>
sin(x)=0, or cos(x)=±1/√2
=>
Note: the interval was probably meant to be: [0,2π)
x=0, x=π for sin(x)=0
or
x=&pi/4, x=3π/4 for cos(x)=1/√2,
x=5π/4 or x=7π4 for cos(x)=-1/√2.
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Afrim, please do not piggy-back on other posts, the often get overlooked.
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To find the solutions to the trigonometric equation sin(2x)cos(x) - sin(x) = 0 in the interval (0, 2π), we can use algebraic techniques and trigonometric identities. Here's how you can proceed:
Step 1: Simplify the equation
sin(2x)cos(x) - sin(x) = 0
Step 2: Apply the double angle identity for sine, which states that sin(2x) = 2sin(x)cos(x). We can substitute sin(2x) = 2sin(x)cos(x) into the equation:
2sin(x)cos(x)cos(x) - sin(x) = 0
2sin(x)cos^2(x) - sin(x) = 0
Step 3: Factor out sin(x) from the expression:
sin(x)(2cos^2(x) - 1) = 0
Now we have two possible cases:
Case 1: sin(x) = 0
If sin(x) = 0, this means x must be a multiple of π because the sine function is equal to zero at these values. In the interval (0, 2π), the solutions for sin(x) = 0 are x = 0 and x = π.
Case 2: 2cos^2(x) - 1 = 0
To solve this equation, isolate the cosine term:
2cos^2(x) - 1 = 0
2cos^2(x) = 1
cos^2(x) = 1/2
Taking the square root of both sides gives us:
cos(x) = ±(1/√2)
To determine the solutions for cos(x) = ±(1/√2) in the interval (0, 2π), we need to find where the cosine function takes these values.
- For cos(x) = 1/√2, we have x = π/4 and x = 7π/4.
- For cos(x) = -1/√2, we have x = 3π/4 and x = 5π/4.
So the solutions for cos(x) = ±(1/√2) in the given interval are x = π/4, 3π/4, 5π/4, and 7π/4.
Therefore, the solutions to the trigonometric equation sin(2x)cos(x) - sin(x) = 0 in the interval (0, 2π) are: x = 0, π/4, 3π/4, 5π/4, and 7π/4.