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Popular Trigonometry >

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

Solution

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

Solution

x = 0.31415 \dots + 2 πn, x = π - 0.31415 \dots + 2 πn, x = - 0.94247 \dots + 2 πn, x = π + 0.94247 \dots + 2 πn

Degrees

x = 1 8^{\circ} + 36 0^{\circ} n, x = 16 2^{\circ} + 36 0^{\circ} n, x = - 5 4^{\circ} + 36 0^{\circ} n, x = 23 4^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

4 u^{2} + 2 u - 1 = 0 : u = \frac{- 1 + 5}{4}, u = - \frac{1 + 5}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 2, c = - 1

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

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

2^{2} - 4 \cdot 4 (- 1) = 25

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

n ab = n a n b

= 5 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 25

u_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 2 + 2 5}{2 \cdot 4}, u_{2} = \frac{- 2 - 2 5}{2 \cdot 4}

u = \frac{- 2 + 2 5}{2 \cdot 4} : \frac{- 1 + 5}{4}

\frac{- 2 + 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 + 2 5}{8}

Factor

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

Rewrite as

= - 2 \cdot 1 + 25

Factor out common term

2

= 2 (- 1 + 5)

= \frac{2 ( - 1 + 5 )}{8}

Cancel the common factor:

2

= \frac{- 1 + 5}{4}

u = \frac{- 2 - 2 5}{2 \cdot 4} : - \frac{1 + 5}{4}

\frac{- 2 - 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{8}

Factor

- 2 - 25 : - 2 (1 + 5)

- 2 - 25

Rewrite as

= - 2 \cdot 1 - 25

Factor out common term

2

= - 2 (1 + 5)

= - \frac{2 ( 1 + 5 )}{8}

Cancel the common factor:

2

= - \frac{1 + 5}{4}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{4}, u = - \frac{1 + 5}{4}

Substitute back

u = sin (x)

sin (x) = \frac{- 1 + 5}{4}, sin (x) = - \frac{1 + 5}{4}

sin (x) = \frac{- 1 + 5}{4}, sin (x) = - \frac{1 + 5}{4}

sin (x) = \frac{- 1 + 5}{4} : x = arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{4}) + 2 πn

sin (x) = \frac{- 1 + 5}{4}

Apply trig inverse properties

sin (x) = \frac{- 1 + 5}{4}

General solutions for

sin (x) = \frac{- 1 + 5}{4}

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

x = arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{4}) + 2 πn

x = arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{4}) + 2 πn

sin (x) = - \frac{1 + 5}{4} : x = arcsin (- \frac{1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{1 + 5}{4}) + 2 πn

sin (x) = - \frac{1 + 5}{4}

Apply trig inverse properties

sin (x) = - \frac{1 + 5}{4}

General solutions for

sin (x) = - \frac{1 + 5}{4}

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

x = arcsin (- \frac{1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{1 + 5}{4}) + 2 πn

x = arcsin (- \frac{1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{1 + 5}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = arcsin (- \frac{1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{1 + 5}{4}) + 2 πn

Show solutions in decimal form

x = 0.31415 \dots + 2 πn, x = π - 0.31415 \dots + 2 πn, x = - 0.94247 \dots + 2 πn, x = π + 0.94247 \dots + 2 πn

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