Trig solving question....?

Question

Find all the solutions, in the interval 0 ≤ x ≤ 2π of the equation

2cos^2x + 1 = 5 sin x


I just don't get how to solve the equation!! Help!! Thanxxxxx

Answer 1

2cos^2(x) + 1 = 5sin(x)

What you want to do is change everything to sines. This is done by using the identity cos^2(x) = 1 - sin^2(x).

2[1 - sin^2(x)] + 1 = 5sin(x)

Expand and simplify.

2 - 2sin^2(x) + 1 = 5sin(x)

3 - 2sin^2(x) = 5sin(x)

Move everything to the right hand side.

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

Factor.

[2sin(x) - 1] [sin(x) + 3] = 0

Equate each factor to 0, and solve for the solutions.

1) 2sin(x) - 1 = 0
2sin(x) = 1
sin(x) = 1/2, which is true when
x = { π/6, 5π/6 }

2) sin(x) + 3 = 0
sin(x) = -3
Note that the range of sin(x) is from -1 to 1, so -3 falls out of the range, and as a result, this will have no solutions.

Therefore, the only solutions of the equation is
x = { π/6, 5π/6 }

Answer 2

2cos^2x + 1 = 5 sin x
2(1-sin^2x) +1 =5sinx
2 -2sin^2x +1 = 5sinx
2sin^2x + 5sin x -3 = 0
(2sinx -1)(sin x+3) = 0
2sinx -1 = 0 --> sinx = 1/2
So x = 30 degrees and 150 degrees
sinx+3 = 0 --> x = -3 which is impossible and is rejected.
So x = pi/6 and 5pi/6 which is 30 degrees and 150 degrees

Answer 3

If you work through it using the cos expansion, then you get sinX = -1/4
and cosX = 4/5

then do cos and sin inverse on your calculator ( set in radians)

so... X = 0.253 radians and 0.644 radians

Answer 4

wut is 2cos^2x
do u mean 2(cos x)^2x