16 sin^2 x = 16 − 8 cos x
Find all solutions of the equation in the interval [0, 2π)
set
sin²(x) = 1-cos²(x)
16(1-cos²(x))=16-8cos(x)
and substitute c=cos(x)
16c²-8c=0
c(2c-1)=0
c=0 or c=1/2
Find all solutions for c=0 in [0,2π) and all solutions for c=1/2 in [0,2π).
Substitute back the solution and check that they are valid.
To solve this trigonometric equation, we can follow these steps:
Step 1: Rearrange the equation
Move all terms to one side of the equation to get zero on the other side:
16 sin^2 x - 16 + 8 cos x = 0
Step 2: Use the trigonometric identity
sin^2 x = 1 - cos^2 x
Replace sin^2 x with 1 - cos^2 x:
16(1 - cos^2 x) - 16 + 8 cos x = 0
Step 3: Simplify the equation
Distribute 16 into the bracket and combine like terms:
16 - 16cos^2 x - 16 + 8cos x = 0
-16cos^2 x + 8cos x = 0
-8cos x(cos x - 1) = 0
Step 4: Solve for cos x
Set each factor equal to zero and solve for cos x:
cos x = 0 or cos x - 1 = 0
For cos x = 0:
Using the unit circle or a calculator, we find that cos x = 0 at x = π/2 and x = 3π/2 in the interval [0, 2π).
For cos x - 1 = 0:
Add 1 to both sides: cos x = 1
Using the unit circle or a calculator, we find that cos x = 1 at x = 0 in the interval [0, 2π).
Step 5: Write down the solutions
The solutions in the interval [0, 2π) are:
x = π/2, 3π/2, and 0
Therefore, the solution to the equation 16 sin^2 x = 16 - 8 cos x in the interval [0, 2π) is x = π/2, 3π/2, and 0.