# math

posted by .

find all solutions on 0 is greater than or equal to theta < 2pi

cos theta= 2sqrt3/3
4-1/2sin theta= (-16+sqrt3)/4
3(sqrt3)= -3cot theta
-2=-1-2sin theta

• math -

2√3/3 = 1.15
cosθ is never greater than 1.
_____________________

If you mean 4-(1/2)sinθ= (-16+√3)/4
sinθ = 4 - √3/4
sorry, sinθ is never greater than 1.

If you mean 4-(1/2sinθ) = (-16+√3)/4
sinθ = 2/(32-√3) = .066
θ = 3.78°
_________________

cotθ = -√3
θ = 30°
__________________

sinθ = 1/2
θ = 30°

I'll let you figure out the other angles less than 2pi

• math -

yeah im lost i have no idea what i did

## Similar Questions

1. ### Calculus

I wanted to confirm that I solved these problems correctly (we had to convert the polar curves to Cartesian equations). 1.rcos(theta)=1 x=1 2.r=2*sin(theta)+2*cos(theta) r^2=2rsin(theta)+2rcos(theta) x^2+y^2=2y+2x (a little unsure …
2. ### Pre Calc.

Use the sum or difference identity to find the exact value of sin255 degrees. My answer: (-sqrt(2)- sqrt(6)) / (4) Find the value of tan (alpha-beta), if cos alpha= -3/5, sin beta= 5/13, 90<alpha<180, and 90<beta<180. My …
3. ### Physics

An object is shot at 20m/s at an angle theta. The object lands 24m away. What is the angle?
4. ### pre calc

find all solutions on 0 is greater than or equal to theta < 2pi cos theta= 2sqrt3/3 4-1/2sin theta= (-16+sqrt3)/4 3(sqrt3)= -3cot theta -2=-1-2sin theta
5. ### Trigonometry

Thank you STEVE!!!!! Suppose that Cos (theta) = 1/square root 2 if 0<equal to theta , pi/2 then sin(theta) = tan (theta) = If 3pi/2 less than equal to theta , 2pi then sin(theta) = tan (theta) = I know the trig functions but I feel …
6. ### trig

If sin theta is equal to 5/13 and theta is an angle in quadrant II find the value of cos theta, sec theta, tan theta, csc theta, cot theta.
7. ### math

knowing that 2(theta) can be written as theta + theta, find the expression for sin2(theta) and cos2(theta). Is it 2sin(theta)cos(theta)?
8. ### calculus

Find circumference of the circle r=2acos theta. s= Int (0 to 2pi) of Sqrt(4a^2cos^2 theta+4a^2sin^2 theta)d theta =Int (0 to 2pi)2a*Int theta d theta =2a(2pi-0)=4a*pi Book shows 2a*pi. Am I wrong somewhere?