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Solve: cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3)
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cosx(tanx + sqroot3) - sqroot3/2(tanx-sqroot3) = What ?
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
tan2x + 2√3 tanx + 3 = 3/4 (tan2x - 2√3 tanx + 3)(tan2x + 1)
Let u = tanx to unclutter things. You end up needing to solve
3u4 - 9√3 u3 + 8u2 - 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
tan2x + 2√3 tanx + 3 = 3/4 (tan2x - 2√3 tanx + 3)(tan2x + 1)
Let u = tanx to unclutter things. You end up needing to solve
3u4 - 9√3 u3 + 8u2 - 14√3 u + 6 = 0
Good luck. Maybe the original problem is missing something.