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

The general solution for 2sin(x)-3cos(x)=2 is x=-2.74680…+2pin,x= pi/2+2pin

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Popular Trigonometry >

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

Solution

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

Solution

x = - 2.74680 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = - 157.38013 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Add

3 cos (x)

to both sides

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

Square both sides

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

Subtract

(2 + 3 cos (x))^{2}

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

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

Simplify

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

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

Add similar elements:

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

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

Group like terms

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

- 4 + 4 = 0

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 12)^{2} = 1 2^{2}

= 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 )} : - \frac{12}{13}

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

Remove parentheses:

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

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

Add 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}

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

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

Remove parentheses:

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

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

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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{12}{13} : x = arccos (- \frac{12}{13}) + 2 πn, x = - arccos (- \frac{12}{13}) + 2 πn

cos (x) = - \frac{12}{13}

Apply trig inverse properties

cos (x) = - \frac{12}{13}

General solutions for

cos (x) = - \frac{12}{13}

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

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

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

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

Combine all the solutions

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

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

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

Check the solution

arccos (- \frac{12}{13}) + 2 πn :

False

arccos (- \frac{12}{13}) + 2 πn

Plug in

n = 1

arccos (- \frac{12}{13}) + 2 π 1

For

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

plug in

x = arccos (- \frac{12}{13}) + 2 π 1

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

Refine

3.53846 \dots = 2

\Rightarrow False

Check the solution

- arccos (- \frac{12}{13}) + 2 πn :

True

- arccos (- \frac{12}{13}) + 2 πn

Plug in

n = 1

- arccos (- \frac{12}{13}) + 2 π 1

For

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

plug in

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

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

Refine

2 = 2

\Rightarrow True

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

plug in

x = \frac{π}{2} + 2 π 1

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

Refine

2 = 2

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

plug in

x = \frac{3 π}{2} + 2 π 1

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

Refine

- 2 = 2

\Rightarrow False

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

Show solutions in decimal form

x = - 2.74680 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)-3cos(x)=2 ?
The general solution for 2sin(x)-3cos(x)=2 is x=-2.74680…+2pin,x= pi/2+2pin