What is the general solution for 2cos(x)+3sin(x)=2 ?

The general solution for 2cos(x)+3sin(x)=2 is x=2pin,x=pi-1.17600…+2pin

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

Solution

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

Solution

x = 2 πn, x = π - 1.17600 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 112.61986 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

3 sin (x)

from both sides

2 cos (x) = 2 - 3 sin (x)

Square both sides

(2 cos (x))^{2} = (2 - 3 sin (x))^{2}

Subtract

(2 - 3 sin (x))^{2}

from both sides

4 cos^{2} (x) - 4 + 12 sin (x) - 9 sin^{2} (x) = 0

Rewrite using trig identities

- 4 + 12 sin (x) + 4 cos^{2} (x) - 9 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 4 + 12 sin (x) + 4 (1 - sin^{2} (x)) - 9 sin^{2} (x)

Simplify

- 4 + 12 sin (x) + 4 (1 - sin^{2} (x)) - 9 sin^{2} (x) : 12 sin (x) - 13 sin^{2} (x)

- 4 + 12 sin (x) + 4 (1 - sin^{2} (x)) - 9 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)

Multiply the numbers:

4 \cdot 1 = 4

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

= - 4 + 12 sin (x) + 4 - 4 sin^{2} (x) - 9 sin^{2} (x)

Simplify

- 4 + 12 sin (x) + 4 - 4 sin^{2} (x) - 9 sin^{2} (x) : 12 sin (x) - 13 sin^{2} (x)

- 4 + 12 sin (x) + 4 - 4 sin^{2} (x) - 9 sin^{2} (x)

Add similar elements:

- 4 sin^{2} (x) - 9 sin^{2} (x) = - 13 sin^{2} (x)

= - 4 + 12 sin (x) + 4 - 13 sin^{2} (x)

Group like terms

= 12 sin (x) - 13 sin^{2} (x) - 4 + 4

- 4 + 4 = 0

= 12 sin (x) - 13 sin^{2} (x)

= 12 sin (x) - 13 sin^{2} (x)

= 12 sin (x) - 13 sin^{2} (x)

12 sin (x) - 13 sin^{2} (x) = 0

Solve by substitution

12 sin (x) - 13 sin^{2} (x) = 0

Let:

sin (x) = u

12 u - 13 u^{2} = 0

12 u - 13 u^{2} = 0 : u = 0, u = \frac{12}{13}

12 u - 13 u^{2} = 0

Write in the standard form

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

- 13 u^{2} + 12 u = 0

Solve with the quadratic formula

- 13 u^{2} + 12 u = 0

Quadratic Equation Formula:

For

a = - 13, b = 12, c = 0

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 13 ) \cdot 0}{2 ( - 13 )}

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 13 ) \cdot 0}{2 ( - 13 )}

1 2^{2} - 4 (- 13) \cdot 0 = 12

1 2^{2} - 4 (- 13) \cdot 0

Apply rule

- (- a) = a

= 1 2^{2} + 4 \cdot 13 \cdot 0

Apply rule

0 \cdot a = 0

= 1 2^{2} + 0

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

= 1 2^{2}

Apply radical rule:

assuming

a \geq 0

= 12

u_{1, 2} = \frac{- 12 \pm 12}{2 ( - 13 )}

Separate the solutions

u_{1} = \frac{- 12 + 12}{2 ( - 13 )}, u_{2} = \frac{- 12 - 12}{2 ( - 13 )}

u = \frac{- 12 + 12}{2 ( - 13 )} : 0

\frac{- 12 + 12}{2 ( - 13 )}

Remove parentheses:

(- a) = - a

= \frac{- 12 + 12}{- 2 \cdot 13}

Add/Subtract the numbers:

- 12 + 12 = 0

= \frac{0}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{0}{- 26}

Apply the fraction rule:

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

= - \frac{0}{26}

Apply rule

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

= - 0

= 0

u = \frac{- 12 - 12}{2 ( - 13 )} : \frac{12}{13}

\frac{- 12 - 12}{2 ( - 13 )}

Remove parentheses:

(- a) = - a

= \frac{- 12 - 12}{- 2 \cdot 13}

Subtract the numbers:

- 12 - 12 = - 24

= \frac{- 24}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{- 24}{- 26}

Apply the fraction rule:

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

= \frac{24}{26}

Cancel the common factor:

2

= \frac{12}{13}

The solutions to the quadratic equation are:

u = 0, u = \frac{12}{13}

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = \frac{12}{13}

sin (x) = 0, sin (x) = \frac{12}{13}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{12}{13} : x = arcsin (\frac{12}{13}) + 2 πn, x = π - arcsin (\frac{12}{13}) + 2 πn

sin (x) = \frac{12}{13}

Apply trig inverse properties

sin (x) = \frac{12}{13}

General solutions for

sin (x) = \frac{12}{13}

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

x = arcsin (\frac{12}{13}) + 2 πn, x = π - arcsin (\frac{12}{13}) + 2 πn

x = arcsin (\frac{12}{13}) + 2 πn, x = π - arcsin (\frac{12}{13}) + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arcsin (\frac{12}{13}) + 2 πn, x = π - arcsin (\frac{12}{13}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

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

plug in

x = 2 π 1

2 cos (2 π 1) + 3 sin (2 π 1) = 2

Refine

2 = 2

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

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

plug in

x = π + 2 π 1

2 cos (π + 2 π 1) + 3 sin (π + 2 π 1) = 2

Refine

- 2 = 2

\Rightarrow False

Check the solution

arcsin (\frac{12}{13}) + 2 πn :

False

arcsin (\frac{12}{13}) + 2 πn

Plug in

n = 1

arcsin (\frac{12}{13}) + 2 π 1

For

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

plug in

x = arcsin (\frac{12}{13}) + 2 π 1

2 cos (arcsin (\frac{12}{13}) + 2 π 1) + 3 sin (arcsin (\frac{12}{13}) + 2 π 1) = 2

Refine

3.53846 \dots = 2

\Rightarrow False

Check the solution

π - arcsin (\frac{12}{13}) + 2 πn :

True

π - arcsin (\frac{12}{13}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{12}{13}) + 2 π 1

For

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

plug in

x = π - arcsin (\frac{12}{13}) + 2 π 1

2 cos (π - arcsin (\frac{12}{13}) + 2 π 1) + 3 sin (π - arcsin (\frac{12}{13}) + 2 π 1) = 2

Refine

2 = 2

\Rightarrow True

x = 2 πn, x = π - arcsin (\frac{12}{13}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = π - 1.17600 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)+3sin(x)=2 ?
The general solution for 2cos(x)+3sin(x)=2 is x=2pin,x=pi-1.17600…+2pin