What is the general solution for arcsin(x)= pi/8 ?

The general solution for arcsin(x)= pi/8 is x=(sqrt(2-\sqrt{2)})/2

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arcsin(x)= pi/8

Solution

arcsin (x) = \frac{π}{8}

Solution

x = \frac{2 - 2}{2}

Solution steps

arcsin (x) = \frac{π}{8}

Apply trig inverse properties

arcsin (x) = \frac{π}{8}

arcsin (x) = a \Rightarrow x = sin (a)

x = sin (\frac{π}{8})

sin (\frac{π}{8}) = \frac{2 - 2}{2}

sin (\frac{π}{8})

Rewrite using trig identities:

\frac{1 - cos ( \frac{π}{4} )}{2}

sin (\frac{π}{8})

Write

sin (\frac{π}{8})

sin (\frac{\frac{π}{4}}{2})

= sin (\frac{\frac{π}{4}}{2})

Use the Half Angle identity:

sin (\frac{θ}{2}) = \frac{1 - cos ( θ )}{2}

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Substitute

θ

with

\frac{θ}{2}

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

Switch sides

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

Divide both sides by

2

sin^{2} (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

sin (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

= \frac{1 - cos ( \frac{π}{4} )}{2}

= \frac{1 - cos ( \frac{π}{4} )}{2}

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{1 - \frac{2}{2}}{2}

Simplify

\frac{1 - \frac{2}{2}}{2} : \frac{2 - 2}{2}

\frac{1 - \frac{2}{2}}{2}

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

\frac{1 - \frac{2}{2}}{2}

Join

1 - \frac{2}{2} : \frac{2 - 2}{2}

1 - \frac{2}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

= \frac{\frac{2 - 2}{2}}{2}

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 - 2}{4}

= \frac{2 - 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 - 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 - 2}{2}

= \frac{2 - 2}{2}

x = \frac{2 - 2}{2}

x = \frac{2 - 2}{2}

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5 tan (x) = 8

3sin(b)+3=5sin(b)+2

3 sin (b) + 3 = 5 sin (b) + 2

8tan^2(θ)+10tan(θ)+10=7

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

3sec(x)-2sqrt(3)=0

3 sec (x) - 23 = 0

2cos(x)-3=-5

2 cos (x) - 3 = - 5

Frequently Asked Questions (FAQ)

What is the general solution for arcsin(x)= pi/8 ?
The general solution for arcsin(x)= pi/8 is x=(sqrt(2-\sqrt{2)})/2