## erin512 Group Title solve the equation on the interval 0≤x≤2pi tanx=0 one year ago one year ago

1. genius12 Group Title

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.

2. erin512 Group Title

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

3. genius12 Group Title

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

4. erin512 Group Title

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

5. genius12 Group Title

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.

6. genius12 Group Title

@erin512

7. ton12 Group Title

tanx = 0 ---> x = n pie

8. ton12 Group Title

Since the intervel is 0≤x≤2pi n = 0,1,2

9. erin512 Group Title

n can equal any number?

10. erin512 Group Title

oh nevermind thank you

11. ton12 Group Title

Welcome

12. attila3453 Group Title

Generally, when tanx = a, x = { arctan(a) + k*pi, when k is whole }

13. erin512 Group Title

so x = 1, 2?

14. erin512 Group Title

i mean x=0,1,2

15. DHASHNI Group Title

$0 \le x \le 2\pi$ take tan on all $\tan 0 \le \tan x \le \tan 2\pi$ => $0 \le tanx \le 0$ that means tanx =0 $\ x=tan^{-1} 0= 0 = 2pi$