What is the general solution for sin^2(θ)-2sin(θ)+1=0 ?

The general solution for sin^2(θ)-2sin(θ)+1=0 is θ= pi/2+2pin

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sin^2(θ)-2sin(θ)+1=0

sin^{2} (θ) - 2 sin (θ) + 1 = 0

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (θ) - 2 sin (θ) + 1 = 0

Solve by substitution

sin^{2} (θ) - 2 sin (θ) + 1 = 0

Let:

sin (θ) = u

u^{2} - 2 u + 1 = 0

u^{2} - 2 u + 1 = 0 : u = 1

u^{2} - 2 u + 1 = 0

Solve with the quadratic formula

u^{2} - 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 1

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

(- 2)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

u = \frac{- ( - 2 )}{2 \cdot 1}

\frac{- ( - 2 )}{2 \cdot 1} = 1

\frac{- ( - 2 )}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = sin (θ)

sin (θ) = 1

sin (θ) = 1

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

General solutions for

sin (θ) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

cos (4 θ) = - \frac{1}{2}

sin (x) = \frac{2 5}{5}

(2 cos (x) - 1) (sin (x) - 1) = 0

cos (3 2^{\circ}) = sin (5 m) - 12

3 = 2 - 2 sin (θ)

What is the general solution for sin^2(θ)-2sin(θ)+1=0 ?
The general solution for sin^2(θ)-2sin(θ)+1=0 is θ= pi/2+2pin