What is the general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 ?

The general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 is x=0.32175…+pin,x=-1.24904…+pin

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6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0

Solution

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x) = 0

Solution

x = 0.32175 \dots + πn, x = - 1.24904 \dots + πn

Degrees

x = 18.43494 \dots^{\circ} + 18 0^{\circ} n, x = - 71.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x) = 0

Factor

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x) : 2 (3 sin (x) - cos (x)) (sin (x) + 3 cos (x))

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x)

Rewrite

- 6

3 \cdot 2

Rewrite

16

8 \cdot 2

= 3 \cdot 2 sin^{2} (x) + 8 \cdot 2 sin (x) cos (x) + 3 \cdot 2 cos^{2} (x)

Factor out common term

2

= 2 (3 sin^{2} (x) + 8 sin (x) cos (x) - 3 cos^{2} (x))

Factor

3 sin^{2} (x) + 8 sin (x) cos (x) - 3 cos^{2} (x) : (3 sin (x) - cos (x)) (sin (x) + 3 cos (x))

3 sin^{2} (x) + 8 sin (x) cos (x) - 3 cos^{2} (x)

Break the expression into groups

3 sin^{2} (x) + 8 sin (x) cos (x) - 3 cos^{2} (x)

Definition

Factors of

9 : 1, 3, 9

9

Divisors (Factors)

Find the Prime factors of

9 : 3, 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

3

is a prime number, therefore no further factorization is possible

= 3 \cdot 3

Add the prime factors:

3

Add 1 and the number

9

itself

1, 9

The factors of

9

1, 3, 9

Negative factors of

9 : - 1, - 3, - 9

Multiply the factors by

- 1

to get the negative factors

- 1, - 3, - 9

For every two factors such that

u * v = - 9,

check if

u + v = 8

Check

u = 1, v = - 9 : u * v = - 9, u + v = - 8 \Rightarrow

FalseCheck

u = 3, v = - 3 : u * v = - 9, u + v = 0 \Rightarrow

False

u = 9, v = - 1

Group into

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

(3 sin^{2} (x) - sin (x) cos (x)) + (9 sin (x) cos (x) - 3 cos^{2} (x))

= (3 sin^{2} (x) - sin (x) cos (x)) + (9 sin (x) cos (x) - 3 cos^{2} (x))

Factor out

sin (x)

from

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

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

Apply exponent rule:

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

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

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

Factor out common term

sin (x)

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

Factor out

3 cos (x)

from

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

9 sin (x) cos (x) - 3 cos^{2} (x)

Apply exponent rule:

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

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

= 9 sin (x) cos (x) - 3 cos (x) cos (x)

Rewrite

9

3 \cdot 3

= 3 \cdot 3 sin (x) cos (x) - 3 cos (x) cos (x)

Factor out common term

3 cos (x)

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

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

Factor out common term

3 sin (x) - cos (x)

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

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

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

Solving each part separately

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

3 sin (x) - cos (x) = 0 : x = arctan (\frac{1}{3}) + πn

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

3 tan (x) - 1 = 0

3 tan (x) - 1 = 0

Move

1

to the right side

3 tan (x) - 1 = 0

Add

1

to both sides

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

Simplify

3 tan (x) = 1

3 tan (x) = 1

Divide both sides by

3

3 tan (x) = 1

Divide both sides by

3

\frac{3 tan ( x )}{3} = \frac{1}{3}

Simplify

tan (x) = \frac{1}{3}

tan (x) = \frac{1}{3}

Apply trig inverse properties

tan (x) = \frac{1}{3}

General solutions for

tan (x) = \frac{1}{3}

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

sin (x) + 3 cos (x) = 0 : x = arctan (- 3) + πn

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) + 3 = 0

tan (x) + 3 = 0

Move

3

to the right side

tan (x) + 3 = 0

Subtract

3

from both sides

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

Simplify

tan (x) = - 3

tan (x) = - 3

Apply trig inverse properties

tan (x) = - 3

General solutions for

tan (x) = - 3

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

x = arctan (- 3) + πn

x = arctan (- 3) + πn

Combine all the solutions

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

Show solutions in decimal form

x = 0.32175 \dots + πn, x = - 1.24904 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 ?
The general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 is x=0.32175…+pin,x=-1.24904…+pin