SOLUTION: Find two solutions of each equation. Give your answers in degrees (0° &#8804; &#952; < 360°) and in radians(0 &#8804; &#952; < 2&#960;). cos &#952; = 1/2

Algebra.Com
Question 859944: Find two solutions of each equation. Give your answers in degrees
(0° ≤ θ < 360°)
and in radians(0 ≤ θ < 2π).
cos θ = 1/2

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Find two solutions of each equation.
Give your answers in degrees
(0° ≤ θ < 360°)
cos(t) = 1/2
t = 60 or t = 300
--------------------------
and in radians(0 ≤ θ < 2π).
cos θ = 1/2
t = pi/3 or t = (5/6)pi
==============================
Cheers,
Stan H.
==============================

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
The adjacent side is
The rotating vector is
The opposite side is ,
so degrees
and
This is with in the 1st quadrant
--------------------------------------
could also be in the 4th quadrant
The adjacent side is
The rotating vector is
The opposite side is ,
degrees
and
---------------------------
In the the other 2 quadrants, 2nd and 3rd,
the cos is negative

RELATED QUESTIONS

Find two solutions of each equation. Give your answers in degrees (0° ≤ θ <... (answered by lwsshak3)
Find two solutions of each equation. Give your answers in degrees (0° ≤ θ <... (answered by lwsshak3)
Find two solutions of each equation. Give your answers in degrees (0° ≤ θ <... (answered by solver91311,Boreal)
Find two solutions of the equation. Give your answers in degrees (0° ≤ θ <... (answered by lwsshak3)
Find values for θ for 0 ≤θ ≤ 2π . (answers in radians):... (answered by lwsshak3)
Determine all of the solutions in the interval 0° ≤ θ < 360°. (Enter your... (answered by lwsshak3)
Find {{{ 2 }}} exact values of θ in degrees ( {{{ 0 }}} ≤ θ < {{{ 360... (answered by ikleyn)
find exact solutions for theta in degrees: √2 sinθ+1=0, from 0 degrees... (answered by stanbon)
Determine all of the solutions in the interval 0° ≤ θ < 360°. (Enter your... (answered by lwsshak3)