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3psilon

sinx + 2cosx = 0 where 0<=x<=2pi, what can x be equal to

  • one year ago
  • one year ago

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  1. nightwill
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    sinx=-2cosx sinx/cosx=-2 tanx=-2 Whatever x that fits tanx=-2

    • one year ago
  2. 3psilon
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    so I did \[\tan^{-1}(-2)\] and I got -1.107 which isn't correct

    • one year ago
  3. 3psilon
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    @nightwill

    • one year ago
  4. nightwill
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    x has to be between 0 and 2pi, so -1.107 won't work.

    • one year ago
  5. nightwill
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    the answer should be pi+arctan(-2) which is about 2.03444

    • one year ago
  6. 3psilon
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    Why did you add pi ?

    • one year ago
  7. 3psilon
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    I'm trying to refresh on trig

    • one year ago
  8. nightwill
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    I added pi so that it fits in the domain. tan x has a period of pi, so tan x = tan (x+pi) Therefore, \[\tan(tan^{-1}-2)=tan(\pi+tan^{-1}-2)\]

    • one year ago
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