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

The general solution for 2cos(x)+1=sin(x) is x= pi/2+2pin,x=-2.49809…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = - 2.49809 \dots + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = - 143.13010 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Square both sides

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

(1 + 2 cos (x))^{2} : 1 + 4 cos (x) + 4 cos^{2} (x)

Apply Perfect Square Formula:

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

a = 1, b = 2 cos (x)

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

Simplify

1^{2} + 2 \cdot 1 \cdot 2 cos (x) + (2 cos (x))^{2} : 1 + 4 cos (x) + 4 cos^{2} (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 2 cos (x) = 4 cos (x)

2 \cdot 1 \cdot 2 cos (x)

Multiply the numbers:

2 \cdot 1 \cdot 2 = 4

= 4 cos (x)

(2 cos (x))^{2} = 4 cos^{2} (x)

(2 cos (x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} cos^{2} (x)

2^{2} = 4

= 4 cos^{2} (x)

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

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

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

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

1 - 1 = 0

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

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

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

4 cos (x) + 5 cos^{2} (x) = 0

Solve by substitution

4 cos (x) + 5 cos^{2} (x) = 0

Let:

cos (x) = u

4 u + 5 u^{2} = 0

4 u + 5 u^{2} = 0 : u = 0, u = - \frac{4}{5}

4 u + 5 u^{2} = 0

Write in the standard form

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

5 u^{2} + 4 u = 0

Solve with the quadratic formula

5 u^{2} + 4 u = 0

Quadratic Equation Formula:

For

a = 5, b = 4, c = 0

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

4^{2} - 4 \cdot 5 \cdot 0 = 4

4^{2} - 4 \cdot 5 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

u_{1, 2} = \frac{- 4 \pm 4}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 4 + 4}{2 \cdot 5}, u_{2} = \frac{- 4 - 4}{2 \cdot 5}

u = \frac{- 4 + 4}{2 \cdot 5} : 0

\frac{- 4 + 4}{2 \cdot 5}

Add/Subtract the numbers:

- 4 + 4 = 0

= \frac{0}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{0}{10}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 4 - 4}{2 \cdot 5} : - \frac{4}{5}

\frac{- 4 - 4}{2 \cdot 5}

Subtract the numbers:

- 4 - 4 = - 8

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 8}{10}

Apply the fraction rule:

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

= - \frac{8}{10}

Cancel the common factor:

2

= - \frac{4}{5}

The solutions to the quadratic equation are:

u = 0, u = - \frac{4}{5}

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = - \frac{4}{5}

cos (x) = 0, cos (x) = - \frac{4}{5}

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

cos (x) = - \frac{4}{5} : x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

cos (x) = - \frac{4}{5}

Apply trig inverse properties

cos (x) = - \frac{4}{5}

General solutions for

cos (x) = - \frac{4}{5}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

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

plug in

x = \frac{π}{2} + 2 π 1

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

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

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

plug in

x = \frac{3 π}{2} + 2 π 1

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

Refine

1 = - 1

\Rightarrow False

Check the solution

arccos (- \frac{4}{5}) + 2 πn :

False

arccos (- \frac{4}{5}) + 2 πn

Plug in

n = 1

arccos (- \frac{4}{5}) + 2 π 1

For

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

plug in

x = arccos (- \frac{4}{5}) + 2 π 1

2 cos (arccos (- \frac{4}{5}) + 2 π 1) + 1 = sin (arccos (- \frac{4}{5}) + 2 π 1)

Refine

- 0.6 = 0.6

\Rightarrow False

Check the solution

- arccos (- \frac{4}{5}) + 2 πn :

True

- arccos (- \frac{4}{5}) + 2 πn

Plug in

n = 1

- arccos (- \frac{4}{5}) + 2 π 1

For

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

plug in

x = - arccos (- \frac{4}{5}) + 2 π 1

2 cos (- arccos (- \frac{4}{5}) + 2 π 1) + 1 = sin (- arccos (- \frac{4}{5}) + 2 π 1)

Refine

- 0.6 = - 0.6

\Rightarrow True

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

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = - 2.49809 \dots + 2 πn

Popular Examples

csc^2(x)=2csc(x)

csc^{2} (x) = 2 csc (x)

cos(x)-3sin(x)=0

cos (x) - 3 sin (x) = 0

2cos(4x)+3=4cos(2x)

2 cos (4 x) + 3 = 4 cos (2 x)

sinh(y)=0

sinh (y) = 0

2+2cos(θ)=4

2 + 2 cos (θ) = 4

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)+1=sin(x) ?
The general solution for 2cos(x)+1=sin(x) is x= pi/2+2pin,x=-2.49809…+2pin