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anonymous
 3 years ago
solve the equation on the interval 0≤x≤2pi
tanx=0
anonymous
 3 years ago
solve the equation on the interval 0≤x≤2pi tanx=0

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Erin have you tried solving this yourself? I encourage you to learn from the solutions I provided and try and solve this yourself first. If you get stuck, then feel free to ask.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you, i'm a bit confused but i'll try solving it myself

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0One hint is that tan(x) = sin(x) / cos(x) and the saying tan(x) = 0 is the same as saying sin(x) = 0. @erin512

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks again, ill try that could you maybe give me an explanation of just one more to help me understand and ill do the next one on my own? i'd really appreciate it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remember what I said. tan(x) = 0 is the same as finding x for sin(x) = 0. This is because tan(x) = sin(x) / cos(x). Since tangent is the quotient of sine and cosine, and we are finding the values of x for where it's 0, we are really finding when the quotient sin(x) / cos(x) is 0. The quotient is 0 when sin(x) is 0 because when the numerator is 0, the whole fraction is 0. Where is sin(x) = 0 ? Look at the sine graph if you need to.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tanx = 0 > x = n pie

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since the intervel is 0≤x≤2pi n = 0,1,2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0n can equal any number?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh nevermind thank you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Generally, when tanx = a, x = { arctan(a) + k*pi, when k is whole }

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[0 \le x \le 2\pi \] take tan on all \[\tan 0 \le \tan x \le \tan 2\pi\] => \[0 \le tanx \le 0 \] that means tanx =0 \[\ x=tan^{1} 0= 0 = 2pi \]
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