What is the general solution for cos(2θ)+sin(2θ)=-1 ?

The general solution for cos(2θ)+sin(2θ)=-1 is θ=pin+pi/2 ,θ=pin+(3pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2θ)+sin(2θ)=-1

Solution

cos (2 θ) + sin (2 θ) = - 1

Solution

θ = πn + \frac{π}{2}, θ = πn + \frac{3 π}{4}

Degrees

θ = 9 0^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 θ) + sin (2 θ) = - 1

Rewrite using trig identities

sin (2 θ) + cos (2 θ)

sin (2 θ) + cos (2 θ) = 2 sin (2 θ + \frac{π}{4})

sin (2 θ) + cos (2 θ)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 θ) + sin (\frac{π}{4}) cos (2 θ))

Use the Angle Sum identity:

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

= 2 sin (2 θ + \frac{π}{4})

= 2 sin (2 θ + \frac{π}{4})

2 sin (2 θ + \frac{π}{4}) = - 1

Divide both sides by

2

2 sin (2 θ + \frac{π}{4}) = - 1

Divide both sides by

2

\frac{2 sin ( 2 θ + \frac{π}{4} )}{2} = \frac{- 1}{2}

Simplify

\frac{2 sin ( 2 θ + \frac{π}{4} )}{2} = \frac{- 1}{2}

Simplify

\frac{2 sin ( 2 θ + \frac{π}{4} )}{2} : sin (2 θ + \frac{π}{4})

\frac{2 sin ( 2 θ + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (2 θ + \frac{π}{4})

Simplify

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

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

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

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (2 θ + \frac{π}{4}) = - \frac{2}{2}

sin (2 θ + \frac{π}{4}) = - \frac{2}{2}

sin (2 θ + \frac{π}{4}) = - \frac{2}{2}

General solutions for

sin (2 θ + \frac{π}{4}) = - \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}

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

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

Solve

2 θ + \frac{π}{4} = \frac{5 π}{4} + 2 πn : θ = πn + \frac{π}{2}

2 θ + \frac{π}{4} = \frac{5 π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

2 θ + \frac{π}{4} = \frac{5 π}{4} + 2 πn

Subtract

\frac{π}{4}

from both sides

2 θ + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{4} + 2 πn - \frac{π}{4}

Simplify

2 θ + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{4} + 2 πn - \frac{π}{4}

Simplify

2 θ + \frac{π}{4} - \frac{π}{4} : 2 θ

2 θ + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= 2 θ

Simplify

\frac{5 π}{4} + 2 πn - \frac{π}{4} : 2 πn + π

\frac{5 π}{4} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{5 π}{4}

Combine the fractions

- \frac{π}{4} + \frac{5 π}{4} : π

Apply rule

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

= \frac{- π + 5 π}{4}

Add similar elements:

- π + 5 π = 4 π

= \frac{4 π}{4}

Divide the numbers:

\frac{4}{4} = 1

= π

= 2 πn + π

2 θ = 2 πn + π

2 θ = 2 πn + π

2 θ = 2 πn + π

Divide both sides by

2

2 θ = 2 πn + π

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{π}{2}

Simplify

θ = πn + \frac{π}{2}

θ = πn + \frac{π}{2}

Solve

2 θ + \frac{π}{4} = \frac{7 π}{4} + 2 πn : θ = πn + \frac{3 π}{4}

2 θ + \frac{π}{4} = \frac{7 π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

2 θ + \frac{π}{4} = \frac{7 π}{4} + 2 πn

Subtract

\frac{π}{4}

from both sides

2 θ + \frac{π}{4} - \frac{π}{4} = \frac{7 π}{4} + 2 πn - \frac{π}{4}

Simplify

2 θ + \frac{π}{4} - \frac{π}{4} = \frac{7 π}{4} + 2 πn - \frac{π}{4}

Simplify

2 θ + \frac{π}{4} - \frac{π}{4} : 2 θ

2 θ + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= 2 θ

Simplify

\frac{7 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{3 π}{2}

\frac{7 π}{4} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{7 π}{4}

Combine the fractions

- \frac{π}{4} + \frac{7 π}{4} : \frac{3 π}{2}

Apply rule

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

= \frac{- π + 7 π}{4}

Add similar elements:

- π + 7 π = 6 π

= \frac{6 π}{4}

Cancel the common factor:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

2 θ = 2 πn + \frac{3 π}{2}

2 θ = 2 πn + \frac{3 π}{2}

2 θ = 2 πn + \frac{3 π}{2}

Divide both sides by

2

2 θ = 2 πn + \frac{3 π}{2}

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

Simplify

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2} : πn + \frac{3 π}{4}

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

= πn + \frac{3 π}{4}

θ = πn + \frac{3 π}{4}

θ = πn + \frac{3 π}{4}

θ = πn + \frac{3 π}{4}

θ = πn + \frac{π}{2}, θ = πn + \frac{3 π}{4}

Graph

Popular Examples

sin(θ)= 19/27

sin (θ) = \frac{19}{27}

sin^4(x)= 1/2 cos^2(x)

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

3cos^2(x)+cos(x)-2=0

3 cos^{2} (x) + cos (x) - 2 = 0

sinh^2(x)+3tanh^2(x)=4

sinh^{2} (x) + 3 tanh^{2} (x) = 4

(sin(51))/(18)=(sin(c))/(17)

\frac{sin ( 5 1 ^{\circ} )}{18} = \frac{sin ( c )}{17}

Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ)+sin(2θ)=-1 ?
The general solution for cos(2θ)+sin(2θ)=-1 is θ=pin+pi/2 ,θ=pin+(3pi)/4