Approximate the equation's solutions in the interval (0,2pi).
sin2x sinx = cosx
I know that I should work on the left side first. I know how to solve for the intervals but I am just not sure how to start this off.
sin2x sinx = cosx
2sinxcosxsinx - cosx = 0
cosx(2sin^2x - 1) = 0
cosx = 0 or sin^2x = 1/2
if cosx = 0, x = pi/2 or 3pi/2
if sin^2x = 1/2
sinx = ± 1/√2
(x could be in all 4 quadrants)
x = pi/4, pi-pi/4, pi+pi/4, and 2pi-pi/4
= ....
(I don't understand why they want you to approximate, we have exact answers)
To approximate the solutions of the equation sin(2x)sin(x) = cos(x) in the interval (0,2pi), we can follow these steps:
Step 1: Simplify the equation
Since we want to work on the left side first, let's simplify sin(2x)sin(x). Using the double angle identity for sin(2x), we get:
sin(2x)sin(x) = (2sin(x)cos(x))sin(x) = 2sin^2(x)cos(x).
Therefore, our equation becomes:
2sin^2(x)cos(x) = cos(x).
Step 2: Rewrite the equation
We can further simplify by dividing both sides of the equation by cos(x):
2sin^2(x) = 1.
Step 3: Solve for sin(x)
Let's solve for sin(x) by taking the square root of both sides:
√(2sin^2(x)) = ±√1.
This simplifies to:
√2|sin(x)| = ±1.
Step 4: Eliminate absolute values
To remove the absolute value, we consider two cases:
Case 1: √2sin(x) = 1
In this case, solve for sin(x):
sin(x) = 1/√2 = √2/2.
Case 2: √2sin(x) = -1
In this case, solve for sin(x):
sin(x) = -1/√2 = -√2/2.
Step 5: Find the angles
Now that we have the values of sin(x), we can find the angles in the interval (0,2pi) that correspond to these values.
For sin(x) = √2/2, we have two possible angles:
x₁ = π/4 (45 degrees)
x₂ = 3π/4 (135 degrees)
For sin(x) = -√2/2, we also have two possible angles:
x₃ = 5π/4 (225 degrees)
x₄ = 7π/4 (315 degrees)
Therefore, the approximate solutions of the equation sin(2x)sin(x) = cos(x) in the interval (0,2pi) are x = π/4, 3π/4, 5π/4, and 7π/4.