What is the general solution for 4sin(x)-3cos(x)=5 ?

The general solution for 4sin(x)-3cos(x)=5 is x=2.21429…+2pin

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4sin(x)-3cos(x)=5

Solution

4 sin (x) - 3 cos (x) = 5

Solution

x = 2.21429 \dots + 2 πn

Degrees

x = 126.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (x) - 3 cos (x) = 5

Add

3 cos (x)

to both sides

4 sin (x) = 5 + 3 cos (x)

Square both sides

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

Subtract

(5 + 3 cos (x))^{2}

from both sides

16 sin^{2} (x) - 25 - 30 cos (x) - 9 cos^{2} (x) = 0

Rewrite using trig identities

- 25 + 16 sin^{2} (x) - 30 cos (x) - 9 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - 25 + 16 (1 - cos^{2} (x)) - 30 cos (x) - 9 cos^{2} (x)

Simplify

- 25 + 16 (1 - cos^{2} (x)) - 30 cos (x) - 9 cos^{2} (x) : - 25 cos^{2} (x) - 30 cos (x) - 9

- 25 + 16 (1 - cos^{2} (x)) - 30 cos (x) - 9 cos^{2} (x)

Expand

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

16 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (x)

= 16 \cdot 1 - 16 cos^{2} (x)

Multiply the numbers:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (x)

= - 25 + 16 - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x)

Simplify

- 25 + 16 - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x) : - 25 cos^{2} (x) - 30 cos (x) - 9

- 25 + 16 - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x)

Group like terms

= - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x) - 25 + 16

Add similar elements:

- 16 cos^{2} (x) - 9 cos^{2} (x) = - 25 cos^{2} (x)

= - 25 cos^{2} (x) - 30 cos (x) - 25 + 16

Add/Subtract the numbers:

- 25 + 16 = - 9

= - 25 cos^{2} (x) - 30 cos (x) - 9

= - 25 cos^{2} (x) - 30 cos (x) - 9

= - 25 cos^{2} (x) - 30 cos (x) - 9

- 9 - 25 cos^{2} (x) - 30 cos (x) = 0

Solve by substitution

- 9 - 25 cos^{2} (x) - 30 cos (x) = 0

Let:

cos (x) = u

- 9 - 25 u^{2} - 30 u = 0

- 9 - 25 u^{2} - 30 u = 0 : u = - \frac{3}{5}

- 9 - 25 u^{2} - 30 u = 0

Write in the standard form

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

- 25 u^{2} - 30 u - 9 = 0

Solve with the quadratic formula

- 25 u^{2} - 30 u - 9 = 0

Quadratic Equation Formula:

For

a = - 25, b = - 30, c = - 9

u_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 ( - 25 ) ( - 9 )}{2 ( - 25 )}

u_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 ( - 25 ) ( - 9 )}{2 ( - 25 )}

(- 30)^{2} - 4 (- 25) (- 9) = 0

(- 30)^{2} - 4 (- 25) (- 9)

Apply rule

- (- a) = a

= (- 30)^{2} - 4 \cdot 25 \cdot 9

Apply exponent rule:

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

n

is even

(- 30)^{2} = 3 0^{2}

= 3 0^{2} - 4 \cdot 25 \cdot 9

Multiply the numbers:

4 \cdot 25 \cdot 9 = 900

= 3 0^{2} - 900

3 0^{2} = 900

= 900 - 900

Subtract the numbers:

900 - 900 = 0

= 0

u_{1, 2} = \frac{- ( - 30 ) \pm 0}{2 ( - 25 )}

u = \frac{- ( - 30 )}{2 ( - 25 )}

\frac{- ( - 30 )}{2 ( - 25 )} = - \frac{3}{5}

\frac{- ( - 30 )}{2 ( - 25 )}

Remove parentheses:

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

= \frac{30}{- 2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{30}{- 50}

Apply the fraction rule:

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

= - \frac{30}{50}

Cancel the common factor:

10

= - \frac{3}{5}

u = - \frac{3}{5}

The solution to the quadratic equation is:

u = - \frac{3}{5}

Substitute back

u = cos (x)

cos (x) = - \frac{3}{5}

cos (x) = - \frac{3}{5}

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

cos (x) = - \frac{3}{5}

Apply trig inverse properties

cos (x) = - \frac{3}{5}

General solutions for

cos (x) = - \frac{3}{5}

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

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

4 sin (x) - 3 cos (x) = 5

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

Check the solution

arccos (- \frac{3}{5}) + 2 πn :

True

arccos (- \frac{3}{5}) + 2 πn

Plug in

n = 1

arccos (- \frac{3}{5}) + 2 π 1

For

4 sin (x) - 3 cos (x) = 5

plug in

x = arccos (- \frac{3}{5}) + 2 π 1

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

Refine

5 = 5

\Rightarrow True

Check the solution

- arccos (- \frac{3}{5}) + 2 πn :

False

- arccos (- \frac{3}{5}) + 2 πn

Plug in

n = 1

- arccos (- \frac{3}{5}) + 2 π 1

For

4 sin (x) - 3 cos (x) = 5

plug in

x = - arccos (- \frac{3}{5}) + 2 π 1

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

Refine

- 1.4 = 5

\Rightarrow False

x = arccos (- \frac{3}{5}) + 2 πn

Show solutions in decimal form

x = 2.21429 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(x)-3cos(x)=5 ?
The general solution for 4sin(x)-3cos(x)=5 is x=2.21429…+2pin