What is the general solution for sin(x)-2cos(x)=1 ?

The general solution for sin(x)-2cos(x)=1 is x=-2.49809…+2pin,x= pi/2+2pin

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sin(x)-2cos(x)=1

Solution

sin (x) - 2 cos (x) = 1

Solution

x = - 2.49809 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = - 143.13010 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) - 2 cos (x) = 1

Add

2 cos (x)

to both sides

sin (x) = 1 + 2 cos (x)

Square both sides

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

Subtract

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

from both sides

sin^{2} (x) - 1 - 4 cos (x) - 4 cos^{2} (x) = 0

Rewrite using trig identities

- 1 + sin^{2} (x) - 4 cos (x) - 4 cos^{2} (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

- 4 cos (x) - 5 cos^{2} (x) = 0

Solve by substitution

- 4 cos (x) - 5 cos^{2} (x) = 0

Let:

cos (x) = u

- 4 u - 5 u^{2} = 0

- 4 u - 5 u^{2} = 0 : u = - \frac{4}{5}, u = 0

- 4 u - 5 u^{2} = 0

Write in the standard form

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

- 5 u^{2} - 4 u = 0

Solve with the quadratic formula

- 5 u^{2} - 4 u = 0

Quadratic Equation Formula:

For

a = - 5, b = - 4, c = 0

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

(- 4)^{2} - 4 (- 5) \cdot 0 = 4

(- 4)^{2} - 4 (- 5) \cdot 0

Apply rule

- (- a) = a

= (- 4)^{2} + 4 \cdot 5 \cdot 0

Apply exponent rule:

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

n

is even

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

= 4^{2} + 4 \cdot 5 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} + 0

4^{2} + 0 = 4^{2}

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

u_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 4}{2 ( - 5 )}, u_{2} = \frac{- ( - 4 ) - 4}{2 ( - 5 )}

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

\frac{- ( - 4 ) + 4}{2 ( - 5 )}

Remove parentheses:

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

= \frac{4 + 4}{- 2 \cdot 5}

Add the numbers:

4 + 4 = 8

= \frac{8}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{8}{- 10}

Apply the fraction rule:

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

= - \frac{8}{10}

Cancel the common factor:

2

= - \frac{4}{5}

u = \frac{- ( - 4 ) - 4}{2 ( - 5 )} : 0

\frac{- ( - 4 ) - 4}{2 ( - 5 )}

Remove parentheses:

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

= \frac{4 - 4}{- 2 \cdot 5}

Subtract the numbers:

4 - 4 = 0

= \frac{0}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{0}{- 10}

Apply the fraction rule:

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

= - \frac{0}{10}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{4}{5}, u = 0

Substitute back

u = cos (x)

cos (x) = - \frac{4}{5}, cos (x) = 0

cos (x) = - \frac{4}{5}, cos (x) = 0

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (x) - 2 cos (x) = 1

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

Check the solution

arccos (- \frac{4}{5}) + 2 πn :

False

arccos (- \frac{4}{5}) + 2 πn

Plug in

n = 1

arccos (- \frac{4}{5}) + 2 π 1

For

sin (x) - 2 cos (x) = 1

plug in

x = arccos (- \frac{4}{5}) + 2 π 1

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

Refine

2.2 = 1

\Rightarrow False

Check the solution

- arccos (- \frac{4}{5}) + 2 πn :

True

- arccos (- \frac{4}{5}) + 2 πn

Plug in

n = 1

- arccos (- \frac{4}{5}) + 2 π 1

For

sin (x) - 2 cos (x) = 1

plug in

x = - arccos (- \frac{4}{5}) + 2 π 1

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

Refine

1 = 1

\Rightarrow True

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

sin (x) - 2 cos (x) = 1

plug in

x = \frac{π}{2} + 2 π 1

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

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

sin (x) - 2 cos (x) = 1

plug in

x = \frac{3 π}{2} + 2 π 1

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

Refine

- 1 = 1

\Rightarrow False

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

Show solutions in decimal form

x = - 2.49809 \dots + 2 πn, x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-2cos(x)=1 ?
The general solution for sin(x)-2cos(x)=1 is x=-2.49809…+2pin,x= pi/2+2pin