How would u solve for Ø in these equations??? - Grade 12, adv functions.

2sin^2 Ø+3sin Ø-2=0

and

tan^2 Ø-1=0

and

1/1+csc Ø =2
????
and Ø has to be in radians?

Since you haven't specified, I will give you answers on [0, 2π) and the general solutions

2sin^2(θ) + 3sin(θ) - 2 = 0

(2sin(θ) - 1)(sin(θ) + 2) = 0

2sin(θ) -1 = 0 -----> sin(θ) = 1/2

sin(θ) = 1/2 when θ = π/6 or θ = 5π/6 -----> Solutions on [0, 2π)

sin(θ) = 1/2 when θ = π/6 + 2πk or θ = 5π/6 + 2πk, where k is any integer -----> General Solution

sin(θ) = -2 -----> NO REAL SOLUTIONS (sin(θ) exists on the interval [-1,1]

tan^2(θ) = 1 -----> √tan^2(θ) = √1 -----> tan(θ) = ±1

tan(θ) = 1 when θ = π/4 or θ = 5π/4 -----> Solutions on [0, 2π)

tan(θ) = -1 when θ = 3π/4 or θ = 7π/4 -----> Solutions on [0, 2π)

tan(θ) = 1 when θ = π/4 + πk, where k is any integer -----> General Solution

tan(θ) = 1 when θ = 3π/4 + πk, where k is any integer -----> General Solution

1/(1+csc(θ)) = 2 -----> 1 = 2(1 + csc(θ)) -----> 1 = 2 + 2csc(θ) -----> -1 = 2csc(θ) -----> csc(θ) = -1/2

csc(θ) = -1/2 -----> NO REAL SOLUTIONS (csc(θ) exists on the interval (-∞,-1] ∪ [1,∞) )

First: Let sin(t) = x and solve like a quadratic.

Second: tan(t) = ±1... consult the unit circle

Third: -1/2 = csc(t) which gives... you tell me.