I need help, I don't remember what to do.
Find all solutions of the equation in the interval [0,2pi].
(2cosè+sqrt3)(cscè+1)=0
Write your answer in radians in terms of pi.
If there is more than one solution, separate them with commas.
zero if either
2 cos e + sqrt 3 = 0
or if
csc e = -1
for the first (in quadrants 2 and 3 for - cos)
cos e = - sqrt 3/2
that in degrees is 180-30 or 180+30
in radians it is pi +/- pi/6
for the second sin is - so quadrants 3 and 4)
1/sin e = -1
sin e = -1
pi/2 and -pi/2
thanks!
To find all solutions of the equation (2cos(θ) + √3)(csc(θ) + 1) = 0 in the interval [0, 2π], we need to solve for θ when the equation equals zero.
First, let's simplify the equation step by step:
1. Distribute: (2cos(θ) + √3)(csc(θ) + 1) = 0
2cos(θ)csc(θ) + 2cos(θ) + √3csc(θ) + √3 = 0
2. Rewrite cos(θ) and csc(θ) in terms of sin(θ):
2(cos(θ)/sin(θ))(1/sin(θ)) + 2cos(θ) + (√3/sin(θ)) + √3 = 0
2(cos(θ)/sin²(θ)) + 2cos(θ) + (√3/sin(θ)) + √3 = 0
3. Common Denominator:
2(cos(θ)/sin²(θ)) + 2cos(θ) + (√3/sin(θ))(sin(θ)/sin(θ)) + √3 = 0
2(cos(θ)/sin²(θ)) + 2cos(θ) + (√3sin(θ)/sin²(θ)) + √3 = 0
4. Combine like terms:
[2cos(θ) + (√3sin(θ))]/sin²(θ) + 2cos(θ) + √3 = 0
5. Multiply through by sin²(θ):
2cos(θ) + (√3sin(θ)) + 2cos(θ)sin²(θ) + √3sin²(θ) = 0
Now we need to solve this equation for θ. You can use the following steps:
1. Combine like terms:
(2cos(θ) + 2cos(θ)sin²(θ)) + (√3sin(θ) + √3sin²(θ)) = 0
2. Factor:
2cos(θ)(1 + sin²(θ)) + √3sin(θ)(1 + sin(θ)) = 0
3. Apply trigonometric identity:
2cos(θ) + √3sin(θ) = 0 [Use: 1 + sin²(θ) = cos²(θ)]
To find the values of θ that satisfy this equation in the interval [0, 2π], you can graph the equations y = 2cos(θ) and y = -√3sin(θ) and find their intersection points. The x-values of these points will give you the solutions for θ in radians in terms of π.
Alternatively, you can use trigonometric identities, like the double-angle or half-angle formulas, to simplify the equation further and solve for θ analytically.