Find the exact values of x if 0<=x<=pi and 2sin^2(x)-sinx-2sinxcosx+cosx=0
To find the exact values of x that satisfy the given equation, we can start by factoring the quadratic equation in terms of sin(x) and cos(x).
Let's rewrite the equation and rearrange terms:
2sin^2(x) - sin(x) - 2sin(x)cos(x) + cos(x) = 0
Rearranging the terms:
2sin^2(x) - 2sin(x)cos(x) - sin(x) + cos(x) = 0
Now, let's factor the equation:
sin(x)(2sin(x) - 2cos(x)) - (2cos(x) - 1) = 0
Further simplifying:
sin(x)(2(sin(x) - cos(x))) - (2cos(x) - 1) = 0
Now, we can factor out sin(x) from the first term:
sin(x)(2sin(x) - 2cos(x)) - (2cos(x) - 1) = 0
sin(x)(2sin(x) - 2cos(x)) + (1 - 2cos(x)) = 0
Taking out a common factor of -1 from the second term:
sin(x)(2sin(x) - 2cos(x)) - (2cos(x) - 1) = 0
sin(x)(2sin(x) - 2cos(x)) - (1 - 2cos(x)) = 0
Now, we can set each factor equal to zero and solve for x:
1) sin(x) = 0
To find the values of x for this condition, we know that sin(x) = 0 when x is an integer multiple of π:
x = 0, π, 2π, ...
2) 2sin(x) - 2cos(x) = 0
Rearranging this equation:
sin(x) = cos(x)
To find the values of x for this condition, we know that sin(x) = cos(x) when x = π/4 and x = 5π/4. Since we are looking for values of x in the range [0, π], we will only consider x = π/4.
Therefore, the exact values of x that satisfy the equation are:
x = 0, π/4, π, 2π
These are the exact values of x within the given range [0, π] that satisfy the equation 2sin^2(x) - sin(x) - 2sin(x)cos(x) + cos(x) = 0.