What is the general solution for 2sin(2θ)=cos(2θ+30) ?

The general solution for 2sin(2θ)=cos(2θ+30) is θ=(0.33347…)/2+(180n)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(2θ)=cos(2θ+30)

Solution

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

Solution

θ = \frac{0.33347 \dots}{2} + \frac{18 0 ^{\circ} n}{2}

Radians

θ = \frac{0.33347 \dots}{2} + \frac{π}{2} n

Solution steps

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

Rewrite using trig identities

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

Rewrite using trig identities

cos (2 θ + 3 0^{\circ})

Use the Angle Sum identity:

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

= cos (2 θ) cos (3 0^{\circ}) - sin (2 θ) sin (3 0^{\circ})

Simplify

cos (2 θ) cos (3 0^{\circ}) - sin (2 θ) sin (3 0^{\circ}) : \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

cos (2 θ) cos (3 0^{\circ}) - sin (2 θ) sin (3 0^{\circ})

Simplify

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2} cos (2 θ) - sin (3 0^{\circ}) sin (2 θ)

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

= \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

2 sin (2 θ) = \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

2 sin (2 θ) = \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

Subtract

\frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

from both sides

\frac{5}{2} sin (2 θ) - \frac{3}{2} cos (2 θ) = 0

Simplify

\frac{5}{2} sin (2 θ) - \frac{3}{2} cos (2 θ) : \frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{2}

\frac{5}{2} sin (2 θ) - \frac{3}{2} cos (2 θ)

Multiply

\frac{5}{2} sin (2 θ) : \frac{5 sin ( 2 θ )}{2}

\frac{5}{2} sin (2 θ)

Multiply fractions:

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

= \frac{5 sin ( 2 θ )}{2}

= \frac{5 sin ( 2 θ )}{2} - \frac{3}{2} cos (2 θ)

Multiply

\frac{3}{2} cos (2 θ) : \frac{3 cos ( 2 θ )}{2}

\frac{3}{2} cos (2 θ)

Multiply fractions:

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

= \frac{3 cos ( 2 θ )}{2}

= \frac{5 sin ( 2 θ )}{2} - \frac{3 cos ( 2 θ )}{2}

Apply rule

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

= \frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{2}

\frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{2} = 0

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

5 sin (2 θ) - 3 cos (2 θ) = 0

Rewrite using trig identities

5 sin (2 θ) - 3 cos (2 θ) = 0

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

\frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{cos ( 2 θ )} = \frac{0}{cos ( 2 θ )}

Simplify

\frac{5 sin ( 2 θ )}{cos ( 2 θ )} - 3 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

5 tan (2 θ) - 3 = 0

5 tan (2 θ) - 3 = 0

Move

3

to the right side

5 tan (2 θ) - 3 = 0

Add

3

to both sides

5 tan (2 θ) - 3 + 3 = 0 + 3

Simplify

5 tan (2 θ) = 3

5 tan (2 θ) = 3

Divide both sides by

5

5 tan (2 θ) = 3

Divide both sides by

5

\frac{5 tan ( 2 θ )}{5} = \frac{3}{5}

Simplify

tan (2 θ) = \frac{3}{5}

tan (2 θ) = \frac{3}{5}

Apply trig inverse properties

tan (2 θ) = \frac{3}{5}

General solutions for

tan (2 θ) = \frac{3}{5}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

Solve

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n : θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

Divide both sides by

2

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Show solutions in decimal form

θ = \frac{0.33347 \dots}{2} + \frac{18 0 ^{\circ} n}{2}

Popular Examples

cos^2(x)= 3/2

cos^{2} (x) = \frac{3}{2}

2sin^2(x)+cos^2(x)+cos(x)-1=0

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

144^2=43.27^2+72^2-2*43.27*72*cos(x)

14 4^{2} = 43.2 7^{2} + 7 2^{2} - 2 \cdot 43.27 \cdot 72 \cdot cos (x)

sin(8x+48)=cos(-4x+26)

sin (8 x + 48) = cos (- 4 x + 26)

3cos(x)+sec(145)=1

3 cos (x) + sec (14 5^{\circ}) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2θ)=cos(2θ+30) ?
The general solution for 2sin(2θ)=cos(2θ+30) is θ=(0.33347…)/2+(180n)/2