What is the general solution for 3cos(x)+sin(x)=1 ?

The general solution for 3cos(x)+sin(x)=1 is x=-0.92729…+2pin,x= pi/2+2pin

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Popular Trigonometry >

3cos(x)+sin(x)=1

Solution

3 cos (x) + sin (x) = 1

Solution

x = - 0.92729 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = - 53.13010 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (x) + sin (x) = 1

Subtract

sin (x)

from both sides

3 cos (x) = 1 - sin (x)

Square both sides

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

Subtract

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

from both sides

9 cos^{2} (x) - 1 + 2 sin (x) - sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 1 - sin^{2} (x) + 2 sin (x) + 9 (1 - sin^{2} (x)) : 2 sin (x) - 10 sin^{2} (x) + 8

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

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

Simplify

- 1 - sin^{2} (x) + 2 sin (x) + 9 - 9 sin^{2} (x) : 2 sin (x) - 10 sin^{2} (x) + 8

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

Group like terms

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

Add similar elements:

- sin^{2} (x) - 9 sin^{2} (x) = - 10 sin^{2} (x)

= - 10 sin^{2} (x) + 2 sin (x) - 1 + 9

Add/Subtract the numbers:

- 1 + 9 = 8

= 2 sin (x) - 10 sin^{2} (x) + 8

= 2 sin (x) - 10 sin^{2} (x) + 8

= 2 sin (x) - 10 sin^{2} (x) + 8

8 - 10 sin^{2} (x) + 2 sin (x) = 0

Solve by substitution

8 - 10 sin^{2} (x) + 2 sin (x) = 0

Let:

sin (x) = u

8 - 10 u^{2} + 2 u = 0

8 - 10 u^{2} + 2 u = 0 : u = - \frac{4}{5}, u = 1

8 - 10 u^{2} + 2 u = 0

Write in the standard form

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

- 10 u^{2} + 2 u + 8 = 0

Solve with the quadratic formula

- 10 u^{2} + 2 u + 8 = 0

Quadratic Equation Formula:

For

a = - 10, b = 2, c = 8

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

2^{2} - 4 (- 10) \cdot 8 = 18

2^{2} - 4 (- 10) \cdot 8

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 10 \cdot 8

Multiply the numbers:

4 \cdot 10 \cdot 8 = 320

= 2^{2} + 320

2^{2} = 4

= 4 + 320

Add the numbers:

4 + 320 = 324

= 324

Factor the number:

324 = 1 8^{2}

= 1 8^{2}

Apply radical rule:

n a^{n} = a

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- 2 \pm 18}{2 ( - 10 )}

Separate the solutions

u_{1} = \frac{- 2 + 18}{2 ( - 10 )}, u_{2} = \frac{- 2 - 18}{2 ( - 10 )}

u = \frac{- 2 + 18}{2 ( - 10 )} : - \frac{4}{5}

\frac{- 2 + 18}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 18}{- 2 \cdot 10}

Add/Subtract the numbers:

- 2 + 18 = 16

= \frac{16}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{16}{- 20}

Apply the fraction rule:

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

= - \frac{16}{20}

Cancel the common factor:

4

= - \frac{4}{5}

u = \frac{- 2 - 18}{2 ( - 10 )} : 1

\frac{- 2 - 18}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 18}{- 2 \cdot 10}

Subtract the numbers:

- 2 - 18 = - 20

= \frac{- 20}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 20}{- 20}

Apply the fraction rule:

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

= \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{4}{5}, u = 1

Substitute back

u = sin (x)

sin (x) = - \frac{4}{5}, sin (x) = 1

sin (x) = - \frac{4}{5}, sin (x) = 1

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 cos (x) + sin (x) = 1

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) + sin (x) = 1

plug in

x = arcsin (- \frac{4}{5}) + 2 π 1

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

Refine

1 = 1

\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) + sin (x) = 1

plug in

x = π + arcsin (\frac{4}{5}) + 2 π 1

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

Refine

- 2.6 = 1

\Rightarrow False

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

3 cos (x) + sin (x) = 1

plug in

x = \frac{π}{2} + 2 π 1

3 cos (\frac{π}{2} + 2 π 1) + sin (\frac{π}{2} + 2 π 1) = 1

Refine

1 = 1

\Rightarrow True

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

Show solutions in decimal form

x = - 0.92729 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(x)+sin(x)=1 ?
The general solution for 3cos(x)+sin(x)=1 is x=-0.92729…+2pin,x= pi/2+2pin