Questions LLC
Login
or
Sign Up
Ask a New Question
Mathematics
Trigonometry
Sine Function
At how many points is sin 0 =1/2 on the interval [-2pi,2pi]?
a) 4 points
b) 2 points
c) 1point
d) 8 points
1 answer
b) 2 points
You can
ask a new question
or
answer this question
.
Related Questions
Find the values of the inverse function in radians.
sin^-1(0.65) a. 0.71+2pi n and -0.71+2pi n b. 0.71+2pi n and -3.85+2pi n c.
Solve for x given the interval of [0,2pi).
cos^2x=2+2sinx I got (3pi)/2 or 270 degrees.. By the way, does the notation [0,2pi)
What are the graphs of y = cos x and y = sec x in the interval from -2pi to 2pi?
The graph of f(x), a trigonometric function, and the graph of g(x) = c intersect at n points over the interval 0 <= x <= 2pi.
determine the solutions for tanxcos^2x-tanx=0 in the interval XE[-2pi,2pi]
Ok, please describe the steps to me that you used to
How do I simplify arcsin (sin 6 pi) given the interval 0 ≤ theta < 2pi
3. find the four angles that define the fourth root of z1=1+ sqrt3*i
z = 2 * (1/2 + i * sqrt(3)/2) z = 2 * (cos(pi/3 + 2pi * k) +
Find all points of extrema on the interval [0, 2pi] if y=x+sinx
(-1.-1+(3pi/2)) (pi, pi) (-1,0) ((3pi/2), 0) none of these I got
Find all solutions to the equation tan(t)=1/tan (t) in the interval 0<t<2pi.
Solve the equation in the interval [0, 2pi]. The
Use De Moivre's Theorem to write the complex number in trigonometric form
(cos(2pi)/7)+ i sin((2pi)/7))^5
