What is the general solution for sin^2(x)-4sin(x)+1=0 ?

The general solution for sin^2(x)-4sin(x)+1=0 is x=0.27126…+2pin,x=pi-0.27126…+2pin

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Popular >

sin^2(x)-4sin(x)+1=0

Solution

sin^{2} (x) - 4 sin (x) + 1 = 0

Solution

x = 0.27126 \dots + 2 πn, x = π - 0.27126 \dots + 2 πn

Degrees

x = 15.54226 \dots^{\circ} + 36 0^{\circ} n, x = 164.45773 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) - 4 sin (x) + 1 = 0

Solve by substitution

sin^{2} (x) - 4 sin (x) + 1 = 0

Let:

sin (x) = u

u^{2} - 4 u + 1 = 0

u^{2} - 4 u + 1 = 0 : u = 2 + 3, u = 2 - 3

u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

Subtract the numbers:

16 - 4 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

u_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

u = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

\frac{- ( - 4 ) + 2 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 + 2 3}{2}

Factor

4 + 23 : 2 (2 + 3)

4 + 23

Rewrite as

= 2 \cdot 2 + 23

Factor out common term

2

= 2 (2 + 3)

= \frac{2 ( 2 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 + 3

u = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

\frac{- ( - 4 ) - 2 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 - 2 3}{2}

Factor

4 - 23 : 2 (2 - 3)

4 - 23

Rewrite as

= 2 \cdot 2 - 23

Factor out common term

2

= 2 (2 - 3)

= \frac{2 ( 2 - 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 - 3

The solutions to the quadratic equation are:

u = 2 + 3, u = 2 - 3

Substitute back

u = sin (x)

sin (x) = 2 + 3, sin (x) = 2 - 3

sin (x) = 2 + 3, sin (x) = 2 - 3

sin (x) = 2 + 3 :

No Solution

sin (x) = 2 + 3

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 2 - 3 : x = arcsin (2 - 3) + 2 πn, x = π - arcsin (2 - 3) + 2 πn

sin (x) = 2 - 3

Apply trig inverse properties

sin (x) = 2 - 3

General solutions for

sin (x) = 2 - 3

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (2 - 3) + 2 πn, x = π - arcsin (2 - 3) + 2 πn

x = arcsin (2 - 3) + 2 πn, x = π - arcsin (2 - 3) + 2 πn

Combine all the solutions

x = arcsin (2 - 3) + 2 πn, x = π - arcsin (2 - 3) + 2 πn

Show solutions in decimal form

x = 0.27126 \dots + 2 πn, x = π - 0.27126 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-4sin(x)+1=0 ?
The general solution for sin^2(x)-4sin(x)+1=0 is x=0.27126…+2pin,x=pi-0.27126…+2pin