What is the general solution for 6=50sin(x)-15cos(x),0<x< pi/2 ?

The general solution for 6=50sin(x)-15cos(x),0<x< pi/2 is x=0.40665…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6=50sin(x)-15cos(x),0<x< pi/2

Solution

6 = 50 sin (x) - 15 cos (x), 0 < x < \frac{π}{2}

Solution

x = 0.40665 \dots

Degrees

x = 23.29935 \dots^{\circ}

Solution steps

6 = 50 sin (x) - 15 cos (x), 0 < x < \frac{π}{2}

Add

15 cos (x)

to both sides

50 sin (x) = 6 + 15 cos (x)

Square both sides

(50 sin (x))^{2} = (6 + 15 cos (x))^{2}

Subtract

(6 + 15 cos (x))^{2}

from both sides

2500 sin^{2} (x) - 36 - 180 cos (x) - 225 cos^{2} (x) = 0

Rewrite using trig identities

- 36 - 180 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)

= - 36 - 180 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x))

Simplify

- 36 - 180 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x)) : - 2725 cos^{2} (x) - 180 cos (x) + 2464

- 36 - 180 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)

= - 36 - 180 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x)

Simplify

- 36 - 180 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x) : - 2725 cos^{2} (x) - 180 cos (x) + 2464

- 36 - 180 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x)

Group like terms

= - 180 cos (x) - 225 cos^{2} (x) - 2500 cos^{2} (x) - 36 + 2500

Add similar elements:

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

= - 180 cos (x) - 2725 cos^{2} (x) - 36 + 2500

Add/Subtract the numbers:

- 36 + 2500 = 2464

= - 2725 cos^{2} (x) - 180 cos (x) + 2464

= - 2725 cos^{2} (x) - 180 cos (x) + 2464

= - 2725 cos^{2} (x) - 180 cos (x) + 2464

2464 - 180 cos (x) - 2725 cos^{2} (x) = 0

Solve by substitution

2464 - 180 cos (x) - 2725 cos^{2} (x) = 0

Let:

cos (x) = u

2464 - 180 u - 2725 u^{2} = 0

2464 - 180 u - 2725 u^{2} = 0 : u = - \frac{2 ( 9 + 5 2689 )}{545}, u = \frac{2 ( 5 2689 - 9 )}{545}

2464 - 180 u - 2725 u^{2} = 0

Write in the standard form

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

- 2725 u^{2} - 180 u + 2464 = 0

Solve with the quadratic formula

- 2725 u^{2} - 180 u + 2464 = 0

Quadratic Equation Formula:

For

a = - 2725, b = - 180, c = 2464

u_{1, 2} = \frac{- ( - 180 ) \pm ( - 180 ) ^{2} - 4 ( - 2725 ) \cdot 2464}{2 ( - 2725 )}

u_{1, 2} = \frac{- ( - 180 ) \pm ( - 180 ) ^{2} - 4 ( - 2725 ) \cdot 2464}{2 ( - 2725 )}

(- 180)^{2} - 4 (- 2725) \cdot 2464 = 1002689

(- 180)^{2} - 4 (- 2725) \cdot 2464

Apply rule

- (- a) = a

= (- 180)^{2} + 4 \cdot 2725 \cdot 2464

Apply exponent rule:

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

n

is even

(- 180)^{2} = 18 0^{2}

= 18 0^{2} + 4 \cdot 2725 \cdot 2464

Multiply the numbers:

4 \cdot 2725 \cdot 2464 = 26857600

= 18 0^{2} + 26857600

18 0^{2} = 32400

= 32400 + 26857600

Add the numbers:

32400 + 26857600 = 26890000

= 26890000

Prime factorization of

26890000 : 2^{4} \cdot 5^{4} \cdot 2689

26890000

= 2^{4} \cdot 5^{4} \cdot 2689

Apply radical rule:

= 2689 2^{4} 5^{4}

Apply radical rule:

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

= 2^{2} 2689 5^{4}

Apply radical rule:

5^{4} = 5^{\frac{4}{2}} = 5^{2}

= 2^{2} \cdot 5^{2} 2689

Refine

= 1002689

u_{1, 2} = \frac{- ( - 180 ) \pm 100 2689}{2 ( - 2725 )}

Separate the solutions

u_{1} = \frac{- ( - 180 ) + 100 2689}{2 ( - 2725 )}, u_{2} = \frac{- ( - 180 ) - 100 2689}{2 ( - 2725 )}

u = \frac{- ( - 180 ) + 100 2689}{2 ( - 2725 )} : - \frac{2 ( 9 + 5 2689 )}{545}

\frac{- ( - 180 ) + 100 2689}{2 ( - 2725 )}

Remove parentheses:

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

= \frac{180 + 100 2689}{- 2 \cdot 2725}

Multiply the numbers:

2 \cdot 2725 = 5450

= \frac{180 + 100 2689}{- 5450}

Apply the fraction rule:

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

= - \frac{180 + 100 2689}{5450}

Cancel

\frac{180 + 100 2689}{5450} : \frac{2 ( 9 + 5 2689 )}{545}

\frac{180 + 100 2689}{5450}

Factor

180 + 1002689 : 20 (9 + 52689)

180 + 1002689

Rewrite as

= 20 \cdot 9 + 20 \cdot 52689

Factor out common term

20

= 20 (9 + 52689)

= \frac{20 ( 9 + 5 2689 )}{5450}

Cancel the common factor:

10

= \frac{2 ( 9 + 5 2689 )}{545}

= - \frac{2 ( 9 + 5 2689 )}{545}

u = \frac{- ( - 180 ) - 100 2689}{2 ( - 2725 )} : \frac{2 ( 5 2689 - 9 )}{545}

\frac{- ( - 180 ) - 100 2689}{2 ( - 2725 )}

Remove parentheses:

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

= \frac{180 - 100 2689}{- 2 \cdot 2725}

Multiply the numbers:

2 \cdot 2725 = 5450

= \frac{180 - 100 2689}{- 5450}

Apply the fraction rule:

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

180 - 1002689 = - (1002689 - 180)

= \frac{100 2689 - 180}{5450}

Factor

1002689 - 180 : 20 (52689 - 9)

1002689 - 180

Rewrite as

= 20 \cdot 52689 - 20 \cdot 9

Factor out common term

20

= 20 (52689 - 9)

= \frac{20 ( 5 2689 - 9 )}{5450}

Cancel the common factor:

10

= \frac{2 ( 5 2689 - 9 )}{545}

The solutions to the quadratic equation are:

u = - \frac{2 ( 9 + 5 2689 )}{545}, u = \frac{2 ( 5 2689 - 9 )}{545}

Substitute back

u = cos (x)

cos (x) = - \frac{2 ( 9 + 5 2689 )}{545}, cos (x) = \frac{2 ( 5 2689 - 9 )}{545}

cos (x) = - \frac{2 ( 9 + 5 2689 )}{545}, cos (x) = \frac{2 ( 5 2689 - 9 )}{545}

cos (x) = - \frac{2 ( 9 + 5 2689 )}{545}, 0 < x < \frac{π}{2} :

No Solution

cos (x) = - \frac{2 ( 9 + 5 2689 )}{545}, 0 < x < \frac{π}{2}

Apply trig inverse properties

cos (x) = - \frac{2 ( 9 + 5 2689 )}{545}

General solutions for

cos (x) = - \frac{2 ( 9 + 5 2689 )}{545}

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

x = arccos (- \frac{2 ( 9 + 5 2689 )}{545}) + 2 πn, x = - arccos (- \frac{2 ( 9 + 5 2689 )}{545}) + 2 πn

x = arccos (- \frac{2 ( 9 + 5 2689 )}{545}) + 2 πn, x = - arccos (- \frac{2 ( 9 + 5 2689 )}{545}) + 2 πn

Solutions for the range

0 < x < \frac{π}{2}

No Solution

cos (x) = \frac{2 ( 5 2689 - 9 )}{545}, 0 < x < \frac{π}{2} : x = arccos (\frac{2 ( 5 2689 - 9 )}{545})

cos (x) = \frac{2 ( 5 2689 - 9 )}{545}, 0 < x < \frac{π}{2}

Apply trig inverse properties

cos (x) = \frac{2 ( 5 2689 - 9 )}{545}

General solutions for

cos (x) = \frac{2 ( 5 2689 - 9 )}{545}

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

x = arccos (\frac{2 ( 5 2689 - 9 )}{545}) + 2 πn, x = 2 π - arccos (\frac{2 ( 5 2689 - 9 )}{545}) + 2 πn

x = arccos (\frac{2 ( 5 2689 - 9 )}{545}) + 2 πn, x = 2 π - arccos (\frac{2 ( 5 2689 - 9 )}{545}) + 2 πn

Solutions for the range

0 < x < \frac{π}{2}

x = arccos (\frac{2 ( 5 2689 - 9 )}{545})

Combine all the solutions

x = arccos (\frac{2 ( 5 2689 - 9 )}{545})

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

50 sin (x) - 15 cos (x) = 6

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

Check the solution

arccos (\frac{2 ( 5 2689 - 9 )}{545}) :

True

arccos (\frac{2 ( 5 2689 - 9 )}{545})

Plug in

n = 1

arccos (\frac{2 ( 5 2689 - 9 )}{545})

For

50 sin (x) - 15 cos (x) = 6

plug in

x = arccos (\frac{2 ( 5 2689 - 9 )}{545})

50 sin (arccos (\frac{2 ( 5 2689 - 9 )}{545})) - 15 cos (arccos (\frac{2 ( 5 2689 - 9 )}{545})) = 6

Refine

6 = 6

\Rightarrow True

x = arccos (\frac{2 ( 5 2689 - 9 )}{545})

Show solutions in decimal form

x = 0.40665 \dots

Popular Examples

solvefor y,x=3sin(y)solve for

y, x = 3 sin (y)

(tan(θ))/(cot(θ))=1

\frac{tan ( θ )}{cot ( θ )} = 1

4sin^2(x)=4

4 sin^{2} (x) = 4

tan(x)=(-4/3 sqrt(3))/(-4)

tan (x) = \frac{- \frac{4}{3} 3}{- 4}

3-2sin(θ)=-3+2sin(θ)

3 - 2 sin (θ) = - 3 + 2 sin (θ)

Frequently Asked Questions (FAQ)

What is the general solution for 6=50sin(x)-15cos(x),0<x< pi/2 ?
The general solution for 6=50sin(x)-15cos(x),0<x< pi/2 is x=0.40665…