What is the general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 ?

The general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 is x=-0.46364…+pin,x=0.58800…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0

Solution

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

Solution

x = - 0.46364 \dots + πn, x = 0.58800 \dots + πn

Degrees

x = - 26.56505 \dots^{\circ} + 18 0^{\circ} n, x = 33.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Factor

6 sin^{2} (x) - sin (x) cos (x) - 2 cos^{2} (x) : (2 sin (x) + cos (x)) (3 sin (x) - 2 cos (x))

6 sin^{2} (x) - sin (x) cos (x) - 2 cos^{2} (x)

Break the expression into groups

6 sin^{2} (x) - sin (x) cos (x) - 2 cos^{2} (x)

Definition

Factors of

12 : 1, 2, 3, 4, 6, 12

12

Divisors (Factors)

Find the Prime factors of

12 : 2, 2, 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

Multiply the prime factors of

12 : 4, 6

2 \cdot 2 = 4

2 \cdot 3 = 6

4, 6

4, 6

Add the prime factors:

2, 3

Add 1 and the number

12

itself

1, 12

The factors of

12

1, 2, 3, 4, 6, 12

Negative factors of

12 : - 1, - 2, - 3, - 4, - 6, - 12

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 4, - 6, - 12

For every two factors such that

u * v = - 12,

check if

u + v = - 1

Check

u = 1, v = - 12 : u * v = - 12, u + v = - 11 \Rightarrow

FalseCheck

u = 2, v = - 6 : u * v = - 12, u + v = - 4 \Rightarrow

False

u = 3, v = - 4

Group into

(a x^{2} + ux y) + (vx y + c y^{2})

(6 sin^{2} (x) + 3 sin (x) cos (x)) + (- 4 sin (x) cos (x) - 2 cos^{2} (x))

= (6 sin^{2} (x) + 3 sin (x) cos (x)) + (- 4 sin (x) cos (x) - 2 cos^{2} (x))

Factor out

3 sin (x)

from

6 sin^{2} (x) + 3 sin (x) cos (x) : 3 sin (x) (2 sin (x) + cos (x))

6 sin^{2} (x) + 3 sin (x) cos (x)

Apply exponent rule:

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

sin^{2} (x) = sin (x) sin (x)

= 6 sin (x) sin (x) + 3 sin (x) cos (x)

Rewrite

6

2 \cdot 3

= 2 \cdot 3 sin (x) sin (x) + 3 sin (x) cos (x)

Factor out common term

3 sin (x)

= 3 sin (x) (2 sin (x) + cos (x))

Factor out

- 2 cos (x)

from

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

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

Apply exponent rule:

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

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

= - 4 sin (x) cos (x) - 2 cos (x) cos (x)

Rewrite

- 4

2 \cdot 2

= 2 \cdot 2 sin (x) cos (x) - 2 cos (x) cos (x)

Factor out common term

- 2 cos (x)

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

= 3 sin (x) (2 sin (x) + cos (x)) - 2 cos (x) (2 sin (x) + cos (x))

Factor out common term

2 sin (x) + cos (x)

= (2 sin (x) + cos (x)) (3 sin (x) - 2 cos (x))

(2 sin (x) + cos (x)) (3 sin (x) - 2 cos (x)) = 0

Solving each part separately

2 sin (x) + cos (x) = 0 or 3 sin (x) - 2 cos (x) = 0

2 sin (x) + cos (x) = 0 : x = arctan (- \frac{1}{2}) + πn

2 sin (x) + cos (x) = 0

Rewrite using trig identities

2 sin (x) + cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{2 sin ( x ) + cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

\frac{2 sin ( x )}{cos ( x )} + 1 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 tan (x) + 1 = 0

2 tan (x) + 1 = 0

Move

1

to the right side

2 tan (x) + 1 = 0

Subtract

1

from both sides

2 tan (x) + 1 - 1 = 0 - 1

Simplify

2 tan (x) = - 1

2 tan (x) = - 1

Divide both sides by

2

2 tan (x) = - 1

Divide both sides by

2

\frac{2 tan ( x )}{2} = \frac{- 1}{2}

Simplify

tan (x) = - \frac{1}{2}

tan (x) = - \frac{1}{2}

Apply trig inverse properties

tan (x) = - \frac{1}{2}

General solutions for

tan (x) = - \frac{1}{2}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{2}) + πn

x = arctan (- \frac{1}{2}) + πn

3 sin (x) - 2 cos (x) = 0 : x = arctan (\frac{2}{3}) + πn

3 sin (x) - 2 cos (x) = 0

Rewrite using trig identities

3 sin (x) - 2 cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

3 tan (x) - 2 = 0

3 tan (x) - 2 = 0

Move

2

to the right side

3 tan (x) - 2 = 0

Add

2

to both sides

3 tan (x) - 2 + 2 = 0 + 2

Simplify

3 tan (x) = 2

3 tan (x) = 2

Divide both sides by

3

3 tan (x) = 2

Divide both sides by

3

\frac{3 tan ( x )}{3} = \frac{2}{3}

Simplify

tan (x) = \frac{2}{3}

tan (x) = \frac{2}{3}

Apply trig inverse properties

tan (x) = \frac{2}{3}

General solutions for

tan (x) = \frac{2}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{2}{3}) + πn

x = arctan (\frac{2}{3}) + πn

Combine all the solutions

x = arctan (- \frac{1}{2}) + πn, x = arctan (\frac{2}{3}) + πn

Show solutions in decimal form

x = - 0.46364 \dots + πn, x = 0.58800 \dots + πn

Graph

Popular Examples

tan(θ)=(20(9.8))/(25)

solvefor t,0=8sin((pit)/(12))+32

2sin^2(x)+sin(x)-1=0,0<= x<2pi

cos(2pix)*2pi+2pi=0

solvefor x,sin(y)=ycos(x)

Frequently Asked Questions (FAQ)

What is the general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 ?
The general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 is x=-0.46364…+pin,x=0.58800…+pin