Question

Solve the equation: -2sin^2(x)+5cos(x)+1=0, where 0°<=x<360°

Answer 1

I manipulated it to \[-2(1-\cos^2x)+5cosx+1=0\] is this correct?

Answer 2

YAh it looks to be right

Answer 3

after that i get to \[2\cos^2x+5cosx-1\] and I'm stuck

Answer 4

Simplifying
-2sin2(x) + 5cos(x) + 1 = 0

Multiply in2s * x
-2in2sx + 5cos(x) + 1 = 0

Multiply cos * x
-2in2sx + 5cosx + 1 = 0

Reorder the terms:
1 + 5cosx + -2in2sx = 0

Solving
1 + 5cosx + -2in2sx = 0

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + 5cosx + -1 + -2in2sx = 0 + -1

Reorder the terms:
1 + -1 + 5cosx + -2in2sx = 0 + -1

Combine like terms: 1 + -1 = 0
0 + 5cosx + -2in2sx = 0 + -1
5cosx + -2in2sx = 0 + -1

Combine like terms: 0 + -1 = -1
5cosx + -2in2sx = -1

Add '2in2sx' to each side of the equation.
5cosx + -2in2sx + 2in2sx = -1 + 2in2sx

Combine like terms: -2in2sx + 2in2sx = 0
5cosx + 0 = -1 + 2in2sx
5cosx = -1 + 2in2sx

Divide each side by '5osx'.
c = -0.2o-1s-1x-1 + 0.4in2o-1

Simplifying
c = -0.2o-1s-1x-1 + 0.4in2o-1

Reorder the terms:
c = 0.4in2o-1 + -0.2o-1s-1x-1

Answer 5

i typed this myself. your welcome

Answer 6

uh they arent variables they're trig functions

Answer 7

I know.

Answer 8

oh, wait.

Answer 9

oops.