MATH SOLVE

3 months ago

Q:
# Please help me solve this. Solve the given equation in the interval [0,2 π]. Note: The answer must be written as a multiple of π. Give exact answers. Do not use decimal numbers. The answer must be an integer or a fraction. Note that π is already provided with the answer so you just have to find the appropriate multiple. E.g. if the answer is π2 you should enter 1/2. If there is more than one answer write them separated by commas.2(sinx)^2−5cosx+1=0

Accepted Solution

A:

sin^2(x) + cos^2(x) = 1

sin^2(x) = 1 - cos^2(x)

Replace sin^2(x) in the equation with 1 - cos^2(x)

2(1 - cos^2(x)) - 5cosx + 1 = 0

2 - 2cos^2(x) - 5cosx + 1 = 0

2cos^2(x) + 5cos x - 3 = 0

Let cosx = y

2y^2 + 5y - 3 = 0

(2y -1)(y + 3) = 0

y = cosx so:

cosx = ½,

cosx = -3 (gives no solutions since cos domain is -1 to 1)

Therefore x = pi/3, (2pi - pi/3) = 5/3pi

sin^2(x) = 1 - cos^2(x)

Replace sin^2(x) in the equation with 1 - cos^2(x)

2(1 - cos^2(x)) - 5cosx + 1 = 0

2 - 2cos^2(x) - 5cosx + 1 = 0

2cos^2(x) + 5cos x - 3 = 0

Let cosx = y

2y^2 + 5y - 3 = 0

(2y -1)(y + 3) = 0

y = cosx so:

cosx = ½,

cosx = -3 (gives no solutions since cos domain is -1 to 1)

Therefore x = pi/3, (2pi - pi/3) = 5/3pi