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

The general solution for 1+sin(x)=(2)(cos(x)) is x=0.64350…+2pin,x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 0.64350 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Square both sides

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

Subtract

(2 cos (x))^{2}

from both sides

(1 + sin (x))^{2} - 4 cos^{2} (x) = 0

Rewrite using trig identities

(1 + sin (x))^{2} - 4 cos^{2} (x)

Use the Pythagorean identity:

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

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

= (1 + sin (x))^{2} - 4 (1 - sin^{2} (x))

Simplify

(1 + sin (x))^{2} - 4 (1 - sin^{2} (x)) : 5 sin^{2} (x) + 2 sin (x) - 3

(1 + sin (x))^{2} - 4 (1 - sin^{2} (x))

(1 + sin (x))^{2} : 1 + 2 sin (x) + sin^{2} (x)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (x)

= 1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Simplify

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x) : 1 + 2 sin (x) + sin^{2} (x)

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

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

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

= 1 + 2 sin (x) + sin^{2} (x) - 4 (1 - sin^{2} (x))

Expand

- 4 (1 - sin^{2} (x)) : - 4 + 4 sin^{2} (x)

- 4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

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

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (x)

= 1 + 2 sin (x) + sin^{2} (x) - 4 + 4 sin^{2} (x)

Simplify

1 + 2 sin (x) + sin^{2} (x) - 4 + 4 sin^{2} (x) : 5 sin^{2} (x) + 2 sin (x) - 3

1 + 2 sin (x) + sin^{2} (x) - 4 + 4 sin^{2} (x)

Group like terms

= 2 sin (x) + sin^{2} (x) + 4 sin^{2} (x) + 1 - 4

Add similar elements:

sin^{2} (x) + 4 sin^{2} (x) = 5 sin^{2} (x)

= 2 sin (x) + 5 sin^{2} (x) + 1 - 4

Add/Subtract the numbers:

1 - 4 = - 3

= 5 sin^{2} (x) + 2 sin (x) - 3

= 5 sin^{2} (x) + 2 sin (x) - 3

= 5 sin^{2} (x) + 2 sin (x) - 3

- 3 + 2 sin (x) + 5 sin^{2} (x) = 0

Solve by substitution

- 3 + 2 sin (x) + 5 sin^{2} (x) = 0

Let:

sin (x) = u

- 3 + 2 u + 5 u^{2} = 0

- 3 + 2 u + 5 u^{2} = 0 : u = \frac{3}{5}, u = - 1

- 3 + 2 u + 5 u^{2} = 0

Write in the standard form

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

5 u^{2} + 2 u - 3 = 0

Solve with the quadratic formula

5 u^{2} + 2 u - 3 = 0

Quadratic Equation Formula:

For

a = 5, b = 2, c = - 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 3) = 8

2^{2} - 4 \cdot 5 (- 3)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

Add the numbers:

4 + 60 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 2 + 8}{2 \cdot 5}, u_{2} = \frac{- 2 - 8}{2 \cdot 5}

u = \frac{- 2 + 8}{2 \cdot 5} : \frac{3}{5}

\frac{- 2 + 8}{2 \cdot 5}

Add/Subtract the numbers:

- 2 + 8 = 6

= \frac{6}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{6}{10}

Cancel the common factor:

2

= \frac{3}{5}

u = \frac{- 2 - 8}{2 \cdot 5} : - 1

\frac{- 2 - 8}{2 \cdot 5}

Subtract the numbers:

- 2 - 8 = - 10

= \frac{- 10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10}{10}

Apply the fraction rule:

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

= - \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{3}{5}, u = - 1

Substitute back

u = sin (x)

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = \frac{3}{5}

Apply trig inverse properties

sin (x) = \frac{3}{5}

General solutions for

sin (x) = \frac{3}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Check the solution

arcsin (\frac{3}{5}) + 2 πn :

True

arcsin (\frac{3}{5}) + 2 πn

Plug in

n = 1

arcsin (\frac{3}{5}) + 2 π 1

For

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

plug in

x = arcsin (\frac{3}{5}) + 2 π 1

1 + sin (arcsin (\frac{3}{5}) + 2 π 1) = 2 cos (arcsin (\frac{3}{5}) + 2 π 1)

Refine

1.6 = 1.6

\Rightarrow True

Check the solution

π - arcsin (\frac{3}{5}) + 2 πn :

False

π - arcsin (\frac{3}{5}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{3}{5}) + 2 π 1

For

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

plug in

x = π - arcsin (\frac{3}{5}) + 2 π 1

1 + sin (π - arcsin (\frac{3}{5}) + 2 π 1) = 2 cos (π - arcsin (\frac{3}{5}) + 2 π 1)

Refine

1.6 = - 1.6

\Rightarrow False

Check the solution

\frac{3 π}{2} + 2 πn :

True

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

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

plug in

x = \frac{3 π}{2} + 2 π 1

1 + sin (\frac{3 π}{2} + 2 π 1) = 2 cos (\frac{3 π}{2} + 2 π 1)

Refine

0 = 0

\Rightarrow True

x = arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

Show solutions in decimal form

x = 0.64350 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

cos(2x)=-0.32

cos (2 x) = - 0.32

sin((3pi)/2-0)=-cos(x)

sin (\frac{3 π}{2} - 0) = - cos (x)

sec(x)=-7/4

sec (x) = - \frac{7}{4}

28cos^2(θ)=7

28 cos^{2} (θ) = 7

3-cos(6x)*6=0

3 - cos (6 x) \cdot 6 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 1+sin(x)=(2)(cos(x)) ?
The general solution for 1+sin(x)=(2)(cos(x)) is x=0.64350…+2pin,x=(3pi)/2+2pin