2sin ² (θ) - 3sin(θ) + 1 = 0?

Answer 1

Just use the quadratic formula

x = [ −b ± √(b² − 4⋅a⋅c) ] ⁄ 2⋅a
Where...
a = 2; b = −3; c = 1; and x = sin θ

sin θ = [ −(−3) ± √( (−3)² − 4⋅(2)⋅(1)) ] ⁄ 2⋅(2)
sin θ = [ 3 ± √( 9 − 8) ] ⁄ 4
sin θ = [ 3 ± √1 ] ⁄ 4
sin θ = [ 3 ± 1 ] ⁄ 4
sin θ = [ 2 or 4 ] ⁄ 4
sin θ = ½ or 1

Then use inverse trigonometric functions
arcsin sin θ = arcsin ½ ... arcsin sin θ = arcsin 1
θ = arcsin ½ ....... ...... ..... θ = arcsin 1
θ = arcsin ½ ....... ...... ..... θ = arcsin 1

θ = π⁄6 & 5·π⁄6 ...... ..... θ = π⁄2

θ = π⁄6 ± 2·π·n
θ = 5·π⁄6 ± 2·π·n
θ = π⁄2 ± 2·π·n
Where 'n' is an integer

Answer 2

This equation is a slick form of a quadratic. To make things clearer, let's substitute u for sin(θ). With this substitution, 2sin ² (θ) - 3sin(θ) + 1 = 0 becomes 2u^2 - 3u + 1 = 0. Now we just have a normal old quadratic that we can solve for u. Factoring this equation gives us (2u-1)(u-1)=0. so u=1 or 1/2. Now if we reverse our substitution for u, sin(θ)=1 or 1/2.

Now to solve for θ...let's apply arcsin to both sides of the equation, giving us arcsin(sin(θ))=arcsin(1). this gives us θ=pi/2 (or in degrees, θ=90). We can find our other value of θ in much the same way. arcsin(sin(θ))=arcsin(1/2), thus θ=pi/6 or 5*pi/6 (in degrees, this is 30 or 150).

So to sum up, for 0<θ<2(pi), θ=pi/6, pi/2, 5*pi/6.
or in degrees... for 0<θ<360, θ=30, 90, 150.

Answer 3

(2 sin θ - 1) (sin θ - 1) = 0
sin θ = 1 / 2 , sin θ = 1
θ = 30° , 150° , 90°

Answer 4

2sin ² (θ) - 3sin(θ) + 1 = 0
=> (2sinθ - 1)(sinθ - 1) = 0
=> sinθ = 1/2 or sinθ = 1
=> θ = kπ +(-1)^k * (π/6), k ∈ Z
OR, θ = kπ +(-1)^k * (π/2), k ∈ Z