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Popular >

8sin^2(x)-6sin(x)+1=0

Solution

8 sin^{2} (x) - 6 sin (x) + 1 = 0

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 165.52248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 sin^{2} (x) - 6 sin (x) + 1 = 0

Solve by substitution

8 sin^{2} (x) - 6 sin (x) + 1 = 0

Let:

sin (x) = u

8 u^{2} - 6 u + 1 = 0

8 u^{2} - 6 u + 1 = 0 : u = \frac{1}{2}, u = \frac{1}{4}

8 u^{2} - 6 u + 1 = 0

Solve with the quadratic formula

8 u^{2} - 6 u + 1 = 0

Quadratic Equation Formula:

For

a = 8, b = - 6, c = 1

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

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

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

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

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

Multiply the numbers:

4 \cdot 8 \cdot 1 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

6 - 2 = 4

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

Multiply the numbers:

2 \cdot 8 = 16

= \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 (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

sin (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}

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn

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