Find all points on the graph of the function f(x) = 2 sin(x) + (sin(x))^2 at which the tangent line is horizontal. Consider the domain x = [0,2π).
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To find the points on the graph of the function where the tangent line is horizontal, we need to find the values of x where the derivative of the function is equal to zero.
Step 1: Find the derivative of f(x):
To find the derivative of f(x) = 2sin(x) + (sin(x))^2, we can use the chain rule.
Let's start with the first term, 2sin(x):
The derivative of sin(x) with respect to x is cos(x).
Multiply this by 2 to get the derivative of 2sin(x) = 2cos(x).
Now let's find the derivative of the second term, (sin(x))^2:
To do this, we can apply the chain rule again.
The derivative of sin(x) with respect to x is cos(x).
Now apply the power rule to (sin(x))^2:
(2)(sin(x))^(2-1)(cos(x)) = 2sin(x)cos(x).
Now, add the derivatives of the two terms together:
f'(x) = 2cos(x) + 2sin(x)cos(x)
Simplify this to f'(x) = 2(cos(x) + sin(x)cos(x)).
Step 2: Set the derivative equal to zero and solve for x:
To find the values of x where the tangent line is horizontal, we set the derivative f'(x) equal to zero:
2(cos(x) + sin(x)cos(x)) = 0
Divide both sides by 2 to simplify the equation:
cos(x) + sin(x)cos(x) = 0
Now, factor out cos(x):
cos(x)(1 + sin(x)) = 0
To find the values of x, we need to solve the two equations:
1) cos(x) = 0,
2) 1 + sin(x) = 0.
1) For cos(x) = 0:
The values of x that satisfy this equation are x = π/2 and x = 3π/2, as cos(x) = 0 at these points on the unit circle.
2) For 1 + sin(x) = 0:
Subtract 1 from both sides to isolate sin(x):
sin(x) = -1
Here, we see that sin(x) = -1 at x = 3π/2 on the unit circle.
So the points on the graph of the function where the tangent line is horizontal are x = π/2 and x = 3π/2.
Please note that we considered the domain x = [0, 2π), so these values fall within that range.