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genius12Best ResponseYou've already chosen the best response.1
Erin 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.
 one year ago

erin512Best ResponseYou've already chosen the best response.0
thank you, i'm a bit confused but i'll try solving it myself
 one year ago

genius12Best ResponseYou've already chosen the best response.1
One hint is that tan(x) = sin(x) / cos(x) and the saying tan(x) = 0 is the same as saying sin(x) = 0. @erin512
 one year ago

erin512Best ResponseYou've already chosen the best response.0
thanks 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
 one year ago

genius12Best ResponseYou've already chosen the best response.1
Remember 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.
 one year ago

ton12Best ResponseYou've already chosen the best response.0
Since the intervel is 0≤x≤2pi n = 0,1,2
 one year ago

erin512Best ResponseYou've already chosen the best response.0
n can equal any number?
 one year ago

erin512Best ResponseYou've already chosen the best response.0
oh nevermind thank you
 one year ago

attila3453Best ResponseYou've already chosen the best response.0
Generally, when tanx = a, x = { arctan(a) + k*pi, when k is whole }
 one year ago

DHASHNIBest 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 \[\ x=tan^{1} 0= 0 = 2pi \]
 one year ago
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