What is the general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 ?

Q: What is the general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 ?

The general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 is θ=105,θ=165,θ=-75,θ=-15,θ=45,θ=-135

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180

Solution

2 cos^{2} (2 θ) + sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

Solution

θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}, θ = 4 5^{\circ}, θ = - 13 5^{\circ}

Radians

θ = \frac{7 π}{12}, θ = \frac{11 π}{12}, θ = - \frac{5 π}{12}, θ = - \frac{π}{12}, θ = \frac{π}{4}, θ = - \frac{3 π}{4}

Solution steps

2 cos^{2} (2 θ) + sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

Subtract

1

from both sides

2 cos^{2} (2 θ) + sin (2 θ) - 1 = 0

Rewrite using trig identities

- 1 + sin (2 θ) + 2 cos^{2} (2 θ)

Use the Pythagorean identity:

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

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

= - 1 + sin (2 θ) + 2 (1 - sin^{2} (2 θ))

Simplify

- 1 + sin (2 θ) + 2 (1 - sin^{2} (2 θ)) : sin (2 θ) - 2 sin^{2} (2 θ) + 1

- 1 + sin (2 θ) + 2 (1 - sin^{2} (2 θ))

Expand

2 (1 - sin^{2} (2 θ)) : 2 - 2 sin^{2} (2 θ)

2 (1 - sin^{2} (2 θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (2 θ)

= 2 \cdot 1 - 2 sin^{2} (2 θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (2 θ)

= - 1 + sin (2 θ) + 2 - 2 sin^{2} (2 θ)

Simplify

- 1 + sin (2 θ) + 2 - 2 sin^{2} (2 θ) : sin (2 θ) - 2 sin^{2} (2 θ) + 1

- 1 + sin (2 θ) + 2 - 2 sin^{2} (2 θ)

Group like terms

= sin (2 θ) - 2 sin^{2} (2 θ) - 1 + 2

Add/Subtract the numbers:

- 1 + 2 = 1

= sin (2 θ) - 2 sin^{2} (2 θ) + 1

= sin (2 θ) - 2 sin^{2} (2 θ) + 1

= sin (2 θ) - 2 sin^{2} (2 θ) + 1

1 + sin (2 θ) - 2 sin^{2} (2 θ) = 0

Solve by substitution

1 + sin (2 θ) - 2 sin^{2} (2 θ) = 0

Let:

sin (2 θ) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 1

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

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

1^{2} - 4 (- 2) \cdot 1 = 3

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (2 θ)

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

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

sin (2 θ) = - \frac{1}{2}, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}

sin (2 θ) = - \frac{1}{2}, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

General solutions for

sin (2 θ) = - \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}

2 θ = 21 0^{\circ} + 36 0^{\circ} n, 2 θ = 33 0^{\circ} + 36 0^{\circ} n

2 θ = 21 0^{\circ} + 36 0^{\circ} n, 2 θ = 33 0^{\circ} + 36 0^{\circ} n

Solve

2 θ = 21 0^{\circ} + 36 0^{\circ} n : θ = 10 5^{\circ} + 18 0^{\circ} n

2 θ = 21 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 21 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{21 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{21 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{21 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 10 5^{\circ} + 18 0^{\circ} n

\frac{21 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{21 0 ^{\circ}}{2} = 10 5^{\circ}

\frac{21 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{126 0 ^{\circ}}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= 10 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 10 5^{\circ} + 18 0^{\circ} n

θ = 10 5^{\circ} + 18 0^{\circ} n

θ = 10 5^{\circ} + 18 0^{\circ} n

θ = 10 5^{\circ} + 18 0^{\circ} n

Solve

2 θ = 33 0^{\circ} + 36 0^{\circ} n : θ = 16 5^{\circ} + 18 0^{\circ} n

2 θ = 33 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 33 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{33 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{33 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{33 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 16 5^{\circ} + 18 0^{\circ} n

\frac{33 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{33 0 ^{\circ}}{2} = 16 5^{\circ}

\frac{33 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{198 0 ^{\circ}}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= 16 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 16 5^{\circ} + 18 0^{\circ} n

θ = 16 5^{\circ} + 18 0^{\circ} n

θ = 16 5^{\circ} + 18 0^{\circ} n

θ = 16 5^{\circ} + 18 0^{\circ} n

θ = 10 5^{\circ} + 18 0^{\circ} n, θ = 16 5^{\circ} + 18 0^{\circ} n

Solutions for the range

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

General solutions for

sin (2 θ) = 1

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}

2 θ = 9 0^{\circ} + 36 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

Solve

2 θ = 9 0^{\circ} + 36 0^{\circ} n : θ = 4 5^{\circ} + 18 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 9 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 4 5^{\circ} + 18 0^{\circ} n

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{9 0 ^{\circ}}{2} = 4 5^{\circ}

\frac{9 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 4 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

Solutions for the range

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 4 5^{\circ}, θ = - 13 5^{\circ}

Combine all the solutions

θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}, θ = 4 5^{\circ}, θ = - 13 5^{\circ}

Popular Examples

3cos(4x)-2= 3/2 sin(8x)-2

3 cos (4 x) - 2 = \frac{3}{2} sin (8 x) - 2

4*cos(x)sin(x)+3*cos(x)=0

4 \cdot cos (x) sin (x) + 3 \cdot cos (x) = 0

(cot(x))(tan(x)-1)=0

(cot (x)) (tan (x) - 1) = 0

10tan(θ)sec(θ)=10cot(θ)csc(θ)

10 tan (θ) sec (θ) = 10 cot (θ) csc (θ)

arctan(x)=((2*1.8))/([(3170+1.7)-1])

arctan (x) = \frac{( 2 \cdot 1.8 )}{[ ( 3170 + 1.7 ) - 1 ]}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 ?
The general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 is θ=105,θ=165,θ=-75,θ=-15,θ=45,θ=-135