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solve the equation on the interval 0≤x≤2pi tanx=0

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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.
thank you, i'm a bit confused but i'll try solving it myself
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

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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
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.
tanx = 0 ---> x = n pie
Since the intervel is 0≤x≤2pi n = 0,1,2
n can equal any number?
oh nevermind thank you
Generally, when tanx = a, x = { arctan(a) + k*pi, when k is whole }
so x = 1, 2?
i mean x=0,1,2
\[0 \le x \le 2\pi \] 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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