Find the exact solutions of the equation in the interval [0, 2π).
(4 sin x)(cos x)=1
By the double-angle formula for sine:
sin(2x) = 2sin(x)cos(x).
So:
4sin(x)cos(x) = 2sin(2x).
The equation now becomes:
4sin(x)cos(x) = 1
==> 2sin(2x) = 1
==> sin(2x) = 1/2.
This solves to get:
2x = π/6 ± 2πk and 2x = 5π/6 ± 2πk, where k is an integer
==> x = π/12 ± πk and x = 5π/12 ± πk.
All solutions that lie on [0, 2π) are given by k = 0 and k = 1. Therefore:
x = π/12, 5π/12, 13π/12, and 17π/12.
Where did the 13π/12 and 17π/12 come from?
the last equation that anonymus was looking at was
sin (2x) = 1/2
so from our basic triangles and the CAST rule
2x = π/6 , 2x = π-π/6 or 5π/6 , ( 30° or 150°)
then
x = π/12 or x = 5π/12
but the period of sin 2x is π
so by adding π to any answer already found we can get a new answer, so
π/12 + π = 13π/12 and 5π/12 = π = 17π/12
OH, alright. I understand now.
Thank you so much for your help!
To find the exact solutions of the equation (4 sin x)(cos x) = 1 in the interval [0, 2π), we can follow these steps:
Step 1: Simplify the equation:
Multiply the terms on the left side of the equation to get:
4 sin x cos x = 1
Step 2: Use a trigonometric identity:
The product of sine and cosine can be written as the half-angle formula: sin(2x) = 2 sin x cos x. Using this identity, we can rewrite the equation as:
2 sin x cos x = 1/2
Step 3: Rewrite the equation:
Divide both sides of the equation by 2 to isolate the sine function:
sin x cos x = 1/4
Step 4: Apply the double-angle formula:
The product of sine and cosine can also be expressed using a double-angle formula: sin(2θ) = 2sinθcosθ. Using this formula, we have:
sin(2x)/2 = 1/4
Step 5: Find the possible values of 2x:
Multiply both sides of the equation by 2 to isolate sin(2x):
sin(2x) = 2/4 = 1/2
Step 6: Find the angle whose sine is 1/2:
The angle whose sine is 1/2 is π/6. So, we have:
2x = π/6
Step 7: Solve for x:
Divide both sides of the equation by 2 to solve for x:
x = π/12
Step 8: Determine the values of x in the interval [0, 2π):
The solution x = π/12 is within the given interval [0, 2π), so it is a valid solution.
Thus, the only solution to the equation (4 sin x)(cos x) = 1 in the interval [0, 2π) is x = π/12.