What is the general solution for 3sin(x)+4cos(x)-5=0 ?

The general solution for 3sin(x)+4cos(x)-5=0 is x=0.64350…+2pin

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Popular >

3sin(x)+4cos(x)-5=0

Solution

3 sin (x) + 4 cos (x) - 5 = 0

Solution

x = 0.64350 \dots + 2 πn

Degrees

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) + 4 cos (x) - 5 = 0

Subtract

4 cos (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

(3 sin (x) - 5)^{2} - 16 cos^{2} (x) = 0

Rewrite using trig identities

(- 5 + 3 sin (x))^{2} - 16 cos^{2} (x)

Use the Pythagorean identity:

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

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

= (- 5 + 3 sin (x))^{2} - 16 (1 - sin^{2} (x))

Simplify

(- 5 + 3 sin (x))^{2} - 16 (1 - sin^{2} (x)) : 25 sin^{2} (x) - 30 sin (x) + 9

(- 5 + 3 sin (x))^{2} - 16 (1 - sin^{2} (x))

(- 5 + 3 sin (x))^{2} : 25 - 30 sin (x) + 9 sin^{2} (x)

Apply Perfect Square Formula:

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

a = - 5, b = 3 sin (x)

= (- 5)^{2} + 2 (- 5) \cdot 3 sin (x) + (3 sin (x))^{2}

Simplify

(- 5)^{2} + 2 (- 5) \cdot 3 sin (x) + (3 sin (x))^{2} : 25 - 30 sin (x) + 9 sin^{2} (x)

(- 5)^{2} + 2 (- 5) \cdot 3 sin (x) + (3 sin (x))^{2}

Remove parentheses:

(- a) = - a

= (- 5)^{2} - 2 \cdot 5 \cdot 3 sin (x) + (3 sin (x))^{2}

(- 5)^{2} = 25

(- 5)^{2}

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2}

5^{2} = 25

= 25

2 \cdot 5 \cdot 3 sin (x) = 30 sin (x)

2 \cdot 5 \cdot 3 sin (x)

Multiply the numbers:

2 \cdot 5 \cdot 3 = 30

= 30 sin (x)

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

(3 sin (x))^{2}

Apply exponent rule:

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

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

3^{2} = 9

= 9 sin^{2} (x)

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

16 \cdot 1 = 16

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

25 - 16 = 9

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

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

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

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

Solve by substitution

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

Let:

sin (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 \cdot 25 \cdot 9}{2 \cdot 25}

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

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

(- 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 \cdot 25}

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 25 = 50

= \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 = sin (x)

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = \frac{3}{5}

Apply trig inverse properties

sin (x) = \frac{3}{5}

General solutions for

sin (x) = \frac{3}{5}

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

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 sin (x) + 4 cos (x) - 5 = 0

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

Check the solution

arcsin (\frac{3}{5}) + 2 πn :

True

arcsin (\frac{3}{5}) + 2 πn

Plug in

n = 1

arcsin (\frac{3}{5}) + 2 π 1

For

3 sin (x) + 4 cos (x) - 5 = 0

plug in

x = arcsin (\frac{3}{5}) + 2 π 1

3 sin (arcsin (\frac{3}{5}) + 2 π 1) + 4 cos (arcsin (\frac{3}{5}) + 2 π 1) - 5 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

π - arcsin (\frac{3}{5}) + 2 πn :

False

π - arcsin (\frac{3}{5}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{3}{5}) + 2 π 1

For

3 sin (x) + 4 cos (x) - 5 = 0

plug in

x = π - arcsin (\frac{3}{5}) + 2 π 1

3 sin (π - arcsin (\frac{3}{5}) + 2 π 1) + 4 cos (π - arcsin (\frac{3}{5}) + 2 π 1) - 5 = 0

Refine

- 6.4 = 0

\Rightarrow False

x = arcsin (\frac{3}{5}) + 2 πn

Show solutions in decimal form

x = 0.64350 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(x)+4cos(x)-5=0 ?
The general solution for 3sin(x)+4cos(x)-5=0 is x=0.64350…+2pin