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

The general solution for 2sin^2(x)=6cos^2(x) is x=(2pi)/3+pin,x= pi/3+pin

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

Solution

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

Solution

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

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

6 cos^{2} (x)

from both sides

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

Factor

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

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

Rewrite

- 6

3 \cdot 2

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

Factor out common term

2

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

Factor

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

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

Rewrite

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

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

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

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(3)^{2} cos^{2} (x) = (3 cos (x))^{2}

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

Solving each part separately

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

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

General solutions for

tan (x) = - 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

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

Add

3

to both sides

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

Simplify

tan (x) = 3

tan (x) = 3

General solutions for

tan (x) = 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)=6cos^2(x) ?
The general solution for 2sin^2(x)=6cos^2(x) is x=(2pi)/3+pin,x= pi/3+pin