What is the general solution for 50sin(x)+15cos(x)=40,0<x<pi ?

The general solution for 50sin(x)+15cos(x)=40,0<x<pi is x=1.97713…,x=0.58154…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

50sin(x)+15cos(x)=40,0<x<pi

Solution

50 sin (x) + 15 cos (x) = 40, 0 < x < π

Solution

x = 1.97713 \dots, x = 0.58154 \dots

Degrees

x = 113.28144 \dots^{\circ}, x = 33.32006 \dots^{\circ}

Solution steps

50 sin (x) + 15 cos (x) = 40, 0 < x < π

Subtract

15 cos (x)

from both sides

50 sin (x) = 40 - 15 cos (x)

Square both sides

(50 sin (x))^{2} = (40 - 15 cos (x))^{2}

Subtract

(40 - 15 cos (x))^{2}

from both sides

2500 sin^{2} (x) - 1600 + 1200 cos (x) - 225 cos^{2} (x) = 0

Rewrite using trig identities

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x))

Simplify

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x)) : 1200 cos (x) - 2725 cos^{2} (x) + 900

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x))

Expand

2500 (1 - cos^{2} (x)) : 2500 - 2500 cos^{2} (x)

2500 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2500, b = 1, c = cos^{2} (x)

= 2500 \cdot 1 - 2500 cos^{2} (x)

Multiply the numbers:

2500 \cdot 1 = 2500

= 2500 - 2500 cos^{2} (x)

= - 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x)

Simplify

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x) : 1200 cos (x) - 2725 cos^{2} (x) + 900

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x)

Group like terms

= 1200 cos (x) - 225 cos^{2} (x) - 2500 cos^{2} (x) - 1600 + 2500

Add similar elements:

- 225 cos^{2} (x) - 2500 cos^{2} (x) = - 2725 cos^{2} (x)

= 1200 cos (x) - 2725 cos^{2} (x) - 1600 + 2500

Add/Subtract the numbers:

- 1600 + 2500 = 900

= 1200 cos (x) - 2725 cos^{2} (x) + 900

= 1200 cos (x) - 2725 cos^{2} (x) + 900

= 1200 cos (x) - 2725 cos^{2} (x) + 900

900 + 1200 cos (x) - 2725 cos^{2} (x) = 0

Solve by substitution

900 + 1200 cos (x) - 2725 cos^{2} (x) = 0

Let:

cos (x) = u

900 + 1200 u - 2725 u^{2} = 0

900 + 1200 u - 2725 u^{2} = 0 : u = - \frac{6 ( 5 5 - 4 )}{109}, u = \frac{6 ( 4 + 5 5 )}{109}

900 + 1200 u - 2725 u^{2} = 0

Write in the standard form

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

- 2725 u^{2} + 1200 u + 900 = 0

Solve with the quadratic formula

- 2725 u^{2} + 1200 u + 900 = 0

Quadratic Equation Formula:

For

a = - 2725, b = 1200, c = 900

u_{1, 2} = \frac{- 1200 \pm 120 0 ^{2} - 4 ( - 2725 ) \cdot 900}{2 ( - 2725 )}

u_{1, 2} = \frac{- 1200 \pm 120 0 ^{2} - 4 ( - 2725 ) \cdot 900}{2 ( - 2725 )}

120 0^{2} - 4 (- 2725) \cdot 900 = 15005

120 0^{2} - 4 (- 2725) \cdot 900

Apply rule

- (- a) = a

= 120 0^{2} + 4 \cdot 2725 \cdot 900

Multiply the numbers:

4 \cdot 2725 \cdot 900 = 9810000

= 120 0^{2} + 9810000

120 0^{2} = 1440000

= 1440000 + 9810000

Add the numbers:

1440000 + 9810000 = 11250000

= 11250000

Prime factorization of

11250000 : 2^{4} \cdot 3^{2} \cdot 5^{7}

11250000

= 5^{7} \cdot 2^{4} \cdot 3^{2}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 5^{6} \cdot 2^{4} \cdot 3^{2} \cdot 5

Apply radical rule:

= 5 2^{4} 3^{2} 5^{6}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 5 3^{2} 5^{6}

Apply radical rule:

3^{2} = 3

= 2^{2} \cdot 35 5^{6}

Apply radical rule:

5^{6} = 5^{\frac{6}{2}} = 5^{3}

= 5^{3} \cdot 2^{2} \cdot 35

Refine

= 15005

u_{1, 2} = \frac{- 1200 \pm 1500 5}{2 ( - 2725 )}

Separate the solutions

u_{1} = \frac{- 1200 + 1500 5}{2 ( - 2725 )}, u_{2} = \frac{- 1200 - 1500 5}{2 ( - 2725 )}

u = \frac{- 1200 + 1500 5}{2 ( - 2725 )} : - \frac{6 ( 5 5 - 4 )}{109}

\frac{- 1200 + 1500 5}{2 ( - 2725 )}

Remove parentheses:

(- a) = - a

= \frac{- 1200 + 1500 5}{- 2 \cdot 2725}

Multiply the numbers:

2 \cdot 2725 = 5450

= \frac{- 1200 + 1500 5}{- 5450}

Apply the fraction rule:

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

= - \frac{- 1200 + 1500 5}{5450}

Cancel

\frac{- 1200 + 1500 5}{5450} : \frac{6 ( 5 5 - 4 )}{109}

\frac{- 1200 + 1500 5}{5450}

Factor

- 1200 + 15005 : 300 (- 4 + 55)

- 1200 + 15005

Rewrite as

= - 300 \cdot 4 + 300 \cdot 55

Factor out common term

300

= 300 (- 4 + 55)

= \frac{300 ( - 4 + 5 5 )}{5450}

Cancel the common factor:

50

= \frac{6 ( 5 5 - 4 )}{109}

= - \frac{6 ( 5 5 - 4 )}{109}

u = \frac{- 1200 - 1500 5}{2 ( - 2725 )} : \frac{6 ( 4 + 5 5 )}{109}

\frac{- 1200 - 1500 5}{2 ( - 2725 )}

Remove parentheses:

(- a) = - a

= \frac{- 1200 - 1500 5}{- 2 \cdot 2725}

Multiply the numbers:

2 \cdot 2725 = 5450

= \frac{- 1200 - 1500 5}{- 5450}

Apply the fraction rule:

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

- 1200 - 15005 = - (1200 + 15005)

= \frac{1200 + 1500 5}{5450}

Factor

1200 + 15005 : 300 (4 + 55)

1200 + 15005

Rewrite as

= 300 \cdot 4 + 300 \cdot 55

Factor out common term

300

= 300 (4 + 55)

= \frac{300 ( 4 + 5 5 )}{5450}

Cancel the common factor:

50

= \frac{6 ( 4 + 5 5 )}{109}

The solutions to the quadratic equation are:

u = - \frac{6 ( 5 5 - 4 )}{109}, u = \frac{6 ( 4 + 5 5 )}{109}

Substitute back

u = cos (x)

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, cos (x) = \frac{6 ( 4 + 5 5 )}{109}

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, cos (x) = \frac{6 ( 4 + 5 5 )}{109}

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, 0 < x < π : x = arccos (- \frac{6 ( 5 5 - 4 )}{109})

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, 0 < x < π

Apply trig inverse properties

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}

General solutions for

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}

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

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn, x = - arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn, x = - arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn

Solutions for the range

0 < x < π

x = arccos (- \frac{6 ( 5 5 - 4 )}{109})

cos (x) = \frac{6 ( 4 + 5 5 )}{109}, 0 < x < π : x = arccos (\frac{6 ( 4 + 5 5 )}{109})

cos (x) = \frac{6 ( 4 + 5 5 )}{109}, 0 < x < π

Apply trig inverse properties

cos (x) = \frac{6 ( 4 + 5 5 )}{109}

General solutions for

cos (x) = \frac{6 ( 4 + 5 5 )}{109}

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

x = arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn, x = 2 π - arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn

x = arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn, x = 2 π - arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn

Solutions for the range

0 < x < π

x = arccos (\frac{6 ( 4 + 5 5 )}{109})

Combine all the solutions

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}), x = arccos (\frac{6 ( 4 + 5 5 )}{109})

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

50 sin (x) + 15 cos (x) = 40

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

Check the solution

arccos (- \frac{6 ( 5 5 - 4 )}{109}) :

True

arccos (- \frac{6 ( 5 5 - 4 )}{109})

Plug in

n = 1

arccos (- \frac{6 ( 5 5 - 4 )}{109})

For

50 sin (x) + 15 cos (x) = 40

plug in

x = arccos (- \frac{6 ( 5 5 - 4 )}{109})

50 sin (arccos (- \frac{6 ( 5 5 - 4 )}{109})) + 15 cos (arccos (- \frac{6 ( 5 5 - 4 )}{109})) = 40

Refine

40 = 40

\Rightarrow True

Check the solution

arccos (\frac{6 ( 4 + 5 5 )}{109}) :

True

arccos (\frac{6 ( 4 + 5 5 )}{109})

Plug in

n = 1

arccos (\frac{6 ( 4 + 5 5 )}{109})

For

50 sin (x) + 15 cos (x) = 40

plug in

x = arccos (\frac{6 ( 4 + 5 5 )}{109})

50 sin (arccos (\frac{6 ( 4 + 5 5 )}{109})) + 15 cos (arccos (\frac{6 ( 4 + 5 5 )}{109})) = 40

Refine

40 = 40

\Rightarrow True

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}), x = arccos (\frac{6 ( 4 + 5 5 )}{109})

Show solutions in decimal form

x = 1.97713 \dots, x = 0.58154 \dots

Graph

Popular Examples

10sin(x)cos(x)=6cos(x)

sin(x)+cos(x)=sec(x)

sin(x)=-0.2761

sin(2y)=0

7sin^2(θ)-15sin(θ)+8=0

Frequently Asked Questions (FAQ)

What is the general solution for 50sin(x)+15cos(x)=40,0<x<pi ?
The general solution for 50sin(x)+15cos(x)=40,0<x<pi is x=1.97713…,x=0.58154…