What is the general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) ?

The general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) is x=(5pi)/4+2pin,x=(7pi)/4+2pin

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Popular Trigonometry >

(sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2)

Solution

\frac{sin ( 2 x )}{sin ( \frac{7 π}{2} - x )} = 2

Solution

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

\frac{sin ( 2 x )}{sin ( \frac{7 π}{2} - x )} = 2

Rewrite using trig identities

\frac{sin ( 2 x )}{sin ( \frac{7 π}{2} - x )} = 2

Rewrite using trig identities

sin (\frac{7 π}{2} - x)

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= sin (\frac{7 π}{2}) cos (x) - cos (\frac{7 π}{2}) sin (x)

Simplify

sin (\frac{7 π}{2}) cos (x) - cos (\frac{7 π}{2}) sin (x) : - cos (x)

sin (\frac{7 π}{2}) cos (x) - cos (\frac{7 π}{2}) sin (x)

sin (\frac{7 π}{2}) cos (x) = - cos (x)

sin (\frac{7 π}{2}) cos (x)

sin (\frac{7 π}{2}) = - 1

sin (\frac{7 π}{2})

sin (\frac{7 π}{2}) = sin (\frac{3 π}{2})

sin (\frac{7 π}{2})

Rewrite

\frac{7 π}{2}

2 π + \frac{3 π}{2}

= sin (2 π + \frac{3 π}{2})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

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

= sin (\frac{3 π}{2})

= sin (\frac{3 π}{2})

Rewrite using trig identities:

sin (π) cos (\frac{π}{2}) + cos (π) sin (\frac{π}{2})

sin (\frac{3 π}{2})

Write

sin (\frac{3 π}{2})

sin (π + \frac{π}{2})

= sin (π + \frac{π}{2})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (π) cos (\frac{π}{2}) + cos (π) sin (\frac{π}{2})

= sin (π) cos (\frac{π}{2}) + cos (π) sin (\frac{π}{2})

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 0

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{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}

= 1

= 0 \cdot 0 + (- 1) \cdot 1

Simplify

= - 1

= - 1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= - cos (x)

= - cos (x) - cos (\frac{7 π}{2}) sin (x)

cos (\frac{7 π}{2}) sin (x) = 0

cos (\frac{7 π}{2}) sin (x)

cos (\frac{7 π}{2}) = 0

cos (\frac{7 π}{2})

cos (\frac{7 π}{2}) = cos (\frac{3 π}{2})

cos (\frac{7 π}{2})

Rewrite

\frac{7 π}{2}

2 π + \frac{3 π}{2}

= cos (2 π + \frac{3 π}{2})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

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

= cos (\frac{3 π}{2})

= cos (\frac{3 π}{2})

Rewrite using trig identities:

cos (π) cos (\frac{π}{2}) - sin (π) sin (\frac{π}{2})

cos (\frac{3 π}{2})

Write

cos (\frac{3 π}{2})

cos (π + \frac{π}{2})

= cos (π + \frac{π}{2})

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (π) cos (\frac{π}{2}) - sin (π) sin (\frac{π}{2})

= cos (π) cos (\frac{π}{2}) - sin (π) sin (\frac{π}{2})

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 0

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{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}

= 1

= (- 1) \cdot 0 - 0 \cdot 1

Simplify

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) - 0

- cos (x) - 0 = - cos (x)

= - cos (x)

= - cos (x)

\frac{sin ( 2 x )}{- cos ( x )} = 2

Simplify

\frac{sin ( 2 x )}{- cos ( x )} : - \frac{sin ( 2 x )}{cos ( x )}

\frac{sin ( 2 x )}{- cos ( x )}

Apply the fraction rule:

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

= - \frac{sin ( 2 x )}{cos ( x )}

- \frac{sin ( 2 x )}{cos ( x )} = 2

- \frac{sin ( 2 x )}{cos ( x )} = 2

Subtract

2

from both sides

- \frac{sin ( 2 x )}{cos ( x )} - 2 = 0

Simplify

- \frac{sin ( 2 x )}{cos ( x )} - 2 : \frac{- sin ( 2 x ) - 2 cos ( x )}{cos ( x )}

- \frac{sin ( 2 x )}{cos ( x )} - 2

Convert element to fraction:

2 = \frac{2 cos ( x )}{cos ( x )}

= - \frac{sin ( 2 x )}{cos ( x )} - \frac{2 cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( 2 x ) - 2 cos ( x )}{cos ( x )}

\frac{- sin ( 2 x ) - 2 cos ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin (2 x) - 2 cos (x) = 0

Rewrite using trig identities

- sin (2 x) - cos (x) 2

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= - 2 sin (x) cos (x) - 2 cos (x)

- cos (x) 2 - 2 cos (x) sin (x) = 0

Factor

- cos (x) 2 - 2 cos (x) sin (x) : - cos (x) (2 + 2 sin (x))

- cos (x) 2 - 2 cos (x) sin (x)

Factor out common term

cos (x)

= - cos (x) (2 + 2 sin (x))

- cos (x) (2 + 2 sin (x)) = 0

Solving each part separately

cos (x) = 0 or 2 + 2 sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 + 2 sin (x) = 0 : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 + 2 sin (x) = 0

Move

2

to the right side

2 + 2 sin (x) = 0

Subtract

2

from both sides

2 + 2 sin (x) - 2 = 0 - 2

Simplify

2 sin (x) = - 2

2 sin (x) = - 2

Divide both sides by

2

2 sin (x) = - 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (x) = - \frac{2}{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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) ?
The general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) is x=(5pi)/4+2pin,x=(7pi)/4+2pin