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8sin^2(2x)-2sin(2x)-1=0

Solution

8 \cdot sin^{2} (2 x) - 2 \cdot sin (2 x) - 1 = 0

Solution

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn, x = - \frac{0.25268 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.25268 \dots}{2} + πn

Degrees

x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n, x = - 7.23875 \dots^{\circ} + 18 0^{\circ} n, x = 97.23875 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (2 x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 8, b = - 2, c = - 1

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 1 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 1 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 1) = 6

(- 2)^{2} - 4 \cdot 8 (- 1)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 1

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 8 \cdot 1

Multiply the numbers:

4 \cdot 8 \cdot 1 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 8}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 8}

u = \frac{- ( - 2 ) + 6}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 2 ) + 6}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{2 + 6}{2 \cdot 8}

Add the numbers:

2 + 6 = 8

= \frac{8}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{16}

Cancel the common factor:

8

= \frac{1}{2}

u = \frac{- ( - 2 ) - 6}{2 \cdot 8} : - \frac{1}{4}

\frac{- ( - 2 ) - 6}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{2 - 6}{2 \cdot 8}

Subtract the numbers:

2 - 6 = - 4

= \frac{- 4}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4}{16}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{16}

Cancel the common factor:

4

= - \frac{1}{4}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - \frac{1}{4}

Substitute back

u = sin (2 x)

sin (2 x) = \frac{1}{2}, sin (2 x) = - \frac{1}{4}

sin (2 x) = \frac{1}{2}, sin (2 x) = - \frac{1}{4}

sin (2 x) = \frac{1}{2} : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

sin (2 x) = \frac{1}{2}

General solutions for

sin (2 x) = \frac{1}{2}

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}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

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

sin (2 x) = - \frac{1}{4}

Apply trig inverse properties

sin (2 x) = - \frac{1}{4}

General solutions for

sin (2 x) = - \frac{1}{4}

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

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

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

Solve

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

2 x = arcsin (- \frac{1}{4}) + 2 πn

Simplify

arcsin (- \frac{1}{4}) + 2 πn : - arcsin (\frac{1}{4}) + 2 πn

arcsin (- \frac{1}{4}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{4}) = - arcsin (\frac{1}{4})

= - arcsin (\frac{1}{4}) + 2 πn

2 x = - arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x = - arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

Solve

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

2 x = π + arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x = π + arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

Combine all the solutions

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn, x = - \frac{arcsin ( \frac{1}{4} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} + πn

Show solutions in decimal form

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn, x = - \frac{0.25268 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.25268 \dots}{2} + πn

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Solution

Solution

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8sin^2(2x)-2sin(2x)-1=0