Asked by Don
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Solve: cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3)
Provide exact answers .. help
Solve: cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3)
Provide exact answers .. help
Answers
Answered by
Anonymous
cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3) = What ?
Answered by
Steve
Well, if you want
cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3) = 0
rearrange things a bit to get
tanx + √3 = √3/2 (tanx - √3)*secx
square both sides to get
tan<sup>2</sup>x + 2√3 tanx + 3 = 3/4 (tan<sup>2</sup>x - 2√3 tanx + 3)(tan<sup>2</sup>x + 1)
Let u = tanx to unclutter things. You end up needing to solve
3u<sup>4</sup> - 9√3 u<sup>3</sup> + 8u<sup>2</sup> - 14√3 u + 6 = 0
Good luck. Maybe the original problem is missing something.
cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3) = 0
rearrange things a bit to get
tanx + √3 = √3/2 (tanx - √3)*secx
square both sides to get
tan<sup>2</sup>x + 2√3 tanx + 3 = 3/4 (tan<sup>2</sup>x - 2√3 tanx + 3)(tan<sup>2</sup>x + 1)
Let u = tanx to unclutter things. You end up needing to solve
3u<sup>4</sup> - 9√3 u<sup>3</sup> + 8u<sup>2</sup> - 14√3 u + 6 = 0
Good luck. Maybe the original problem is missing something.
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