What is the general solution for 2sin(x)-cos(x)=0.6 ?

The general solution for 2sin(x)-cos(x)=0.6 is x=-2.94960…+2pin,x=0.73530…+2pin

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Popular Trigonometry >

2sin(x)-cos(x)=0.6

Solution

2 sin (x) - cos (x) = 0.6

Solution

x = - 2.94960 \dots + 2 πn, x = 0.73530 \dots + 2 πn

Degrees

x = - 168.99975 \dots^{\circ} + 36 0^{\circ} n, x = 42.12985 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) - cos (x) = 0.6

Add

cos (x)

to both sides

2 sin (x) = 0.6 + cos (x)

Square both sides

(2 sin (x))^{2} = (0.6 + cos (x))^{2}

Subtract

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

from both sides

4 sin^{2} (x) - 0.36 - 1.2 cos (x) - cos^{2} (x) = 0

Rewrite using trig identities

- 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 (1 - cos^{2} (x))

Simplify

- 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 (1 - cos^{2} (x)) : - 5 cos^{2} (x) - 1.2 cos (x) + 3.64

- 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 (1 - cos^{2} (x))

Expand

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

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

= - 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 - 4 cos^{2} (x)

Simplify

- 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 - 4 cos^{2} (x) : - 5 cos^{2} (x) - 1.2 cos (x) + 3.64

- 0.36 - cos^{2} (x) - 1.2 cos (x) + 4 - 4 cos^{2} (x)

Group like terms

= - cos^{2} (x) - 1.2 cos (x) - 4 cos^{2} (x) - 0.36 + 4

Add similar elements:

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

= - 5 cos^{2} (x) - 1.2 cos (x) - 0.36 + 4

Add/Subtract the numbers:

- 0.36 + 4 = 3.64

= - 5 cos^{2} (x) - 1.2 cos (x) + 3.64

= - 5 cos^{2} (x) - 1.2 cos (x) + 3.64

= - 5 cos^{2} (x) - 1.2 cos (x) + 3.64

3.64 - 1.2 cos (x) - 5 cos^{2} (x) = 0

Solve by substitution

3.64 - 1.2 cos (x) - 5 cos^{2} (x) = 0

Let:

cos (x) = u

3.64 - 1.2 u - 5 u^{2} = 0

3.64 - 1.2 u - 5 u^{2} = 0 : u = - \frac{3 + 4 29}{25}, u = \frac{4 29 - 3}{25}

3.64 - 1.2 u - 5 u^{2} = 0

Multiply both sides by

100

3.64 - 1.2 u - 5 u^{2} = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

3.64 \cdot 100 - 1.2 u \cdot 100 - 5 u^{2} \cdot 100 = 0 \cdot 100

Refine

364 - 120 u - 500 u^{2} = 0

364 - 120 u - 500 u^{2} = 0

Write in the standard form

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

- 500 u^{2} - 120 u + 364 = 0

Solve with the quadratic formula

- 500 u^{2} - 120 u + 364 = 0

Quadratic Equation Formula:

For

a = - 500, b = - 120, c = 364

u_{1, 2} = \frac{- ( - 120 ) \pm ( - 120 ) ^{2} - 4 ( - 500 ) \cdot 364}{2 ( - 500 )}

u_{1, 2} = \frac{- ( - 120 ) \pm ( - 120 ) ^{2} - 4 ( - 500 ) \cdot 364}{2 ( - 500 )}

(- 120)^{2} - 4 (- 500) \cdot 364 = 16029

(- 120)^{2} - 4 (- 500) \cdot 364

Apply rule

- (- a) = a

= (- 120)^{2} + 4 \cdot 500 \cdot 364

Apply exponent rule:

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

n

is even

(- 120)^{2} = 12 0^{2}

= 12 0^{2} + 4 \cdot 500 \cdot 364

Multiply the numbers:

4 \cdot 500 \cdot 364 = 728000

= 12 0^{2} + 728000

12 0^{2} = 14400

= 14400 + 728000

Add the numbers:

14400 + 728000 = 742400

= 742400

Prime factorization of

742400 : 2^{10} \cdot 5^{2} \cdot 29

742400

= 2^{10} \cdot 5^{2} \cdot 29

Apply radical rule:

= 29 5^{2} 2^{10}

Apply radical rule:

2^{10} = 2^{\frac{10}{2}} = 2^{5}

= 2^{5} 29 5^{2}

Apply radical rule:

5^{2} = 5

= 2^{5} \cdot 529

Refine

= 16029

u_{1, 2} = \frac{- ( - 120 ) \pm 160 29}{2 ( - 500 )}

Separate the solutions

u_{1} = \frac{- ( - 120 ) + 160 29}{2 ( - 500 )}, u_{2} = \frac{- ( - 120 ) - 160 29}{2 ( - 500 )}

u = \frac{- ( - 120 ) + 160 29}{2 ( - 500 )} : - \frac{3 + 4 29}{25}

\frac{- ( - 120 ) + 160 29}{2 ( - 500 )}

Remove parentheses:

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

= \frac{120 + 160 29}{- 2 \cdot 500}

Multiply the numbers:

2 \cdot 500 = 1000

= \frac{120 + 160 29}{- 1000}

Apply the fraction rule:

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

= - \frac{120 + 160 29}{1000}

Cancel

\frac{120 + 160 29}{1000} : \frac{3 + 4 29}{25}

\frac{120 + 160 29}{1000}

Factor

120 + 16029 : 40 (3 + 429)

120 + 16029

Rewrite as

= 40 \cdot 3 + 40 \cdot 429

Factor out common term

40

= 40 (3 + 429)

= \frac{40 ( 3 + 4 29 )}{1000}

Cancel the common factor:

40

= \frac{3 + 4 29}{25}

= - \frac{3 + 4 29}{25}

u = \frac{- ( - 120 ) - 160 29}{2 ( - 500 )} : \frac{4 29 - 3}{25}

\frac{- ( - 120 ) - 160 29}{2 ( - 500 )}

Remove parentheses:

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

= \frac{120 - 160 29}{- 2 \cdot 500}

Multiply the numbers:

2 \cdot 500 = 1000

= \frac{120 - 160 29}{- 1000}

Apply the fraction rule:

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

120 - 16029 = - (16029 - 120)

= \frac{160 29 - 120}{1000}

Factor

16029 - 120 : 40 (429 - 3)

16029 - 120

Rewrite as

= 40 \cdot 429 - 40 \cdot 3

Factor out common term

40

= 40 (429 - 3)

= \frac{40 ( 4 29 - 3 )}{1000}

Cancel the common factor:

40

= \frac{4 29 - 3}{25}

The solutions to the quadratic equation are:

u = - \frac{3 + 4 29}{25}, u = \frac{4 29 - 3}{25}

Substitute back

u = cos (x)

cos (x) = - \frac{3 + 4 29}{25}, cos (x) = \frac{4 29 - 3}{25}

cos (x) = - \frac{3 + 4 29}{25}, cos (x) = \frac{4 29 - 3}{25}

cos (x) = - \frac{3 + 4 29}{25} : x = arccos (- \frac{3 + 4 29}{25}) + 2 πn, x = - arccos (- \frac{3 + 4 29}{25}) + 2 πn

cos (x) = - \frac{3 + 4 29}{25}

Apply trig inverse properties

cos (x) = - \frac{3 + 4 29}{25}

General solutions for

cos (x) = - \frac{3 + 4 29}{25}

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

x = arccos (- \frac{3 + 4 29}{25}) + 2 πn, x = - arccos (- \frac{3 + 4 29}{25}) + 2 πn

x = arccos (- \frac{3 + 4 29}{25}) + 2 πn, x = - arccos (- \frac{3 + 4 29}{25}) + 2 πn

cos (x) = \frac{4 29 - 3}{25} : x = arccos (\frac{4 29 - 3}{25}) + 2 πn, x = 2 π - arccos (\frac{4 29 - 3}{25}) + 2 πn

cos (x) = \frac{4 29 - 3}{25}

Apply trig inverse properties

cos (x) = \frac{4 29 - 3}{25}

General solutions for

cos (x) = \frac{4 29 - 3}{25}

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

x = arccos (\frac{4 29 - 3}{25}) + 2 πn, x = 2 π - arccos (\frac{4 29 - 3}{25}) + 2 πn

x = arccos (\frac{4 29 - 3}{25}) + 2 πn, x = 2 π - arccos (\frac{4 29 - 3}{25}) + 2 πn

Combine all the solutions

x = arccos (- \frac{3 + 4 29}{25}) + 2 πn, x = - arccos (- \frac{3 + 4 29}{25}) + 2 πn, x = arccos (\frac{4 29 - 3}{25}) + 2 πn, x = 2 π - arccos (\frac{4 29 - 3}{25}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sin (x) - cos (x) = 0.6

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

Check the solution

arccos (- \frac{3 + 4 29}{25}) + 2 πn :

False

arccos (- \frac{3 + 4 29}{25}) + 2 πn

Plug in

n = 1

arccos (- \frac{3 + 4 29}{25}) + 2 π 1

For

2 sin (x) - cos (x) = 0.6

plug in

x = arccos (- \frac{3 + 4 29}{25}) + 2 π 1

2 sin (arccos (- \frac{3 + 4 29}{25}) + 2 π 1) - cos (arccos (- \frac{3 + 4 29}{25}) + 2 π 1) = 0.6

Refine

1.36325 \dots = 0.6

\Rightarrow False

Check the solution

- arccos (- \frac{3 + 4 29}{25}) + 2 πn :

True

- arccos (- \frac{3 + 4 29}{25}) + 2 πn

Plug in

n = 1

- arccos (- \frac{3 + 4 29}{25}) + 2 π 1

For

2 sin (x) - cos (x) = 0.6

plug in

x = - arccos (- \frac{3 + 4 29}{25}) + 2 π 1

2 sin (- arccos (- \frac{3 + 4 29}{25}) + 2 π 1) - cos (- arccos (- \frac{3 + 4 29}{25}) + 2 π 1) = 0.6

Refine

0.6 = 0.6

\Rightarrow True

Check the solution

arccos (\frac{4 29 - 3}{25}) + 2 πn :

True

arccos (\frac{4 29 - 3}{25}) + 2 πn

Plug in

n = 1

arccos (\frac{4 29 - 3}{25}) + 2 π 1

For

2 sin (x) - cos (x) = 0.6

plug in

x = arccos (\frac{4 29 - 3}{25}) + 2 π 1

2 sin (arccos (\frac{4 29 - 3}{25}) + 2 π 1) - cos (arccos (\frac{4 29 - 3}{25}) + 2 π 1) = 0.6

Refine

0.6 = 0.6

\Rightarrow True

Check the solution

2 π - arccos (\frac{4 29 - 3}{25}) + 2 πn :

False

2 π - arccos (\frac{4 29 - 3}{25}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{4 29 - 3}{25}) + 2 π 1

For

2 sin (x) - cos (x) = 0.6

plug in

x = 2 π - arccos (\frac{4 29 - 3}{25}) + 2 π 1

2 sin (2 π - arccos (\frac{4 29 - 3}{25}) + 2 π 1) - cos (2 π - arccos (\frac{4 29 - 3}{25}) + 2 π 1) = 0.6

Refine

- 2.08325 \dots = 0.6

\Rightarrow False

x = - arccos (- \frac{3 + 4 29}{25}) + 2 πn, x = arccos (\frac{4 29 - 3}{25}) + 2 πn

Show solutions in decimal form

x = - 2.94960 \dots + 2 πn, x = 0.73530 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)-cos(x)=0.6 ?
The general solution for 2sin(x)-cos(x)=0.6 is x=-2.94960…+2pin,x=0.73530…+2pin