anonymous
  • anonymous
Solve the equation: -2sin^2(x)+5cos(x)+1=0, where 0°<=x<360°
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I manipulated it to \[-2(1-\cos^2x)+5cosx+1=0\] is this correct?
JesseReyna
  • JesseReyna
YAh it looks to be right
anonymous
  • anonymous
after that i get to \[2\cos^2x+5cosx-1\] and I'm stuck

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Landon34
  • Landon34
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
Landon34
  • Landon34
i typed this myself. your welcome
anonymous
  • anonymous
uh they arent variables they're trig functions
Landon34
  • Landon34
I know.
Landon34
  • Landon34
oh, wait.
Landon34
  • Landon34
oops.

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