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sinx + 2cosx = 0 where 0<=x<=2pi, what can x be equal to
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sinx=-2cosx sinx/cosx=-2 tanx=-2 Whatever x that fits tanx=-2
so I did \[\tan^{-1}(-2)\] and I got -1.107 which isn't correct
@nightwill
x has to be between 0 and 2pi, so -1.107 won't work.
the answer should be pi+arctan(-2) which is about 2.03444
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Why did you add pi ?
I'm trying to refresh on trig
I added pi so that it fits in the domain. tan x has a period of pi, so tan x = tan (x+pi) Therefore, \[\tan(tan^{-1}-2)=tan(\pi+tan^{-1}-2)\]
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