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

The general solution for 3cos(x)-4sin(x)=5 is x=-0.92729…+2pin

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

Solution

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

Solution

x = - 0.92729 \dots + 2 πn

Degrees

x = - 53.13010 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Add

4 sin (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

9 cos^{2} (x) - 25 - 40 sin (x) - 16 sin^{2} (x) = 0

Rewrite using trig identities

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - 25 - 16 sin^{2} (x) - 40 sin (x) + 9 (1 - sin^{2} (x))

Simplify

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 (1 - sin^{2} (x)) : - 25 sin^{2} (x) - 40 sin (x) - 16

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 (1 - sin^{2} (x))

Expand

9 (1 - sin^{2} (x)) : 9 - 9 sin^{2} (x)

9 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = sin^{2} (x)

= 9 \cdot 1 - 9 sin^{2} (x)

Multiply the numbers:

9 \cdot 1 = 9

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

= - 25 - 16 sin^{2} (x) - 40 sin (x) + 9 - 9 sin^{2} (x)

Simplify

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 - 9 sin^{2} (x) : - 25 sin^{2} (x) - 40 sin (x) - 16

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 - 9 sin^{2} (x)

Group like terms

= - 16 sin^{2} (x) - 40 sin (x) - 9 sin^{2} (x) - 25 + 9

Add similar elements:

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

= - 25 sin^{2} (x) - 40 sin (x) - 25 + 9

Add/Subtract the numbers:

- 25 + 9 = - 16

= - 25 sin^{2} (x) - 40 sin (x) - 16

= - 25 sin^{2} (x) - 40 sin (x) - 16

= - 25 sin^{2} (x) - 40 sin (x) - 16

- 16 - 25 sin^{2} (x) - 40 sin (x) = 0

Solve by substitution

- 16 - 25 sin^{2} (x) - 40 sin (x) = 0

Let:

sin (x) = u

- 16 - 25 u^{2} - 40 u = 0

- 16 - 25 u^{2} - 40 u = 0 : u = - \frac{4}{5}

- 16 - 25 u^{2} - 40 u = 0

Write in the standard form

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

- 25 u^{2} - 40 u - 16 = 0

Solve with the quadratic formula

- 25 u^{2} - 40 u - 16 = 0

Quadratic Equation Formula:

For

a = - 25, b = - 40, c = - 16

u_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 25 ) ( - 16 )}{2 ( - 25 )}

u_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 25 ) ( - 16 )}{2 ( - 25 )}

(- 40)^{2} - 4 (- 25) (- 16) = 0

(- 40)^{2} - 4 (- 25) (- 16)

Apply rule

- (- a) = a

= (- 40)^{2} - 4 \cdot 25 \cdot 16

Apply exponent rule:

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

n

is even

(- 40)^{2} = 4 0^{2}

= 4 0^{2} - 4 \cdot 25 \cdot 16

Multiply the numbers:

4 \cdot 25 \cdot 16 = 1600

= 4 0^{2} - 1600

4 0^{2} = 1600

= 1600 - 1600

Subtract the numbers:

1600 - 1600 = 0

= 0

u_{1, 2} = \frac{- ( - 40 ) \pm 0}{2 ( - 25 )}

u = \frac{- ( - 40 )}{2 ( - 25 )}

\frac{- ( - 40 )}{2 ( - 25 )} = - \frac{4}{5}

\frac{- ( - 40 )}{2 ( - 25 )}

Remove parentheses:

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

= \frac{40}{- 2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{40}{- 50}

Apply the fraction rule:

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

= - \frac{40}{50}

Cancel the common factor:

10

= - \frac{4}{5}

u = - \frac{4}{5}

The solution to the quadratic equation is:

u = - \frac{4}{5}

Substitute back

u = sin (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

arcsin (- \frac{4}{5}) + 2 πn :

True

arcsin (- \frac{4}{5}) + 2 πn

Plug in

n = 1

arcsin (- \frac{4}{5}) + 2 π 1

For

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

plug in

x = arcsin (- \frac{4}{5}) + 2 π 1

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

Refine

5 = 5

\Rightarrow True

Check the solution

π + arcsin (\frac{4}{5}) + 2 πn :

False

π + arcsin (\frac{4}{5}) + 2 πn

Plug in

n = 1

π + arcsin (\frac{4}{5}) + 2 π 1

For

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

plug in

x = π + arcsin (\frac{4}{5}) + 2 π 1

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

Refine

1.4 = 5

\Rightarrow False

x = arcsin (- \frac{4}{5}) + 2 πn

Show solutions in decimal form

x = - 0.92729 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(x)-4sin(x)=5 ?
The general solution for 3cos(x)-4sin(x)=5 is x=-0.92729…+2pin