What is the general solution for sin^2(2x)=3cos^2(2x) ?

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

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

Solution

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

Solution

x = \frac{π}{3} + \frac{πn}{2}, x = \frac{π}{6} + \frac{πn}{2}

Degrees

x = 6 0^{\circ} + 9 0^{\circ} n, x = 3 0^{\circ} + 9 0^{\circ} n

Solution steps

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

Subtract

3 cos^{2} (2 x)

from both sides

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

Factor

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

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

Rewrite

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

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

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

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

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

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

Rewrite using trig identities

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

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

tan (2 x) + 3 = 0

tan (2 x) + 3 = 0

Move

3

to the right side

tan (2 x) + 3 = 0

Subtract

3

from both sides

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

Simplify

tan (2 x) = - 3

tan (2 x) = - 3

General solutions for

tan (2 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}

2 x = \frac{2 π}{3} + πn

2 x = \frac{2 π}{3} + πn

Solve

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

2 x = \frac{2 π}{3} + πn

Divide both sides by

2

2 x = \frac{2 π}{3} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{2} + \frac{πn}{2} : \frac{π}{3} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{2} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

= \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

sin (2 x) - 3 cos (2 x) = 0 : x = \frac{π}{6} + \frac{πn}{2}

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

Rewrite using trig identities

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

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

tan (2 x) - 3 = 0

tan (2 x) - 3 = 0

Move

3

to the right side

tan (2 x) - 3 = 0

Add

3

to both sides

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

Simplify

tan (2 x) = 3

tan (2 x) = 3

General solutions for

tan (2 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}

2 x = \frac{π}{3} + πn

2 x = \frac{π}{3} + πn

Solve

2 x = \frac{π}{3} + πn : x = \frac{π}{6} + \frac{πn}{2}

2 x = \frac{π}{3} + πn

Divide both sides by

2

2 x = \frac{π}{3} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{3}}{2} + \frac{πn}{2} : \frac{π}{6} + \frac{πn}{2}

\frac{\frac{π}{3}}{2} + \frac{πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

Combine all the solutions

x = \frac{π}{3} + \frac{πn}{2}, x = \frac{π}{6} + \frac{πn}{2}

Graph

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Frequently Asked Questions (FAQ)

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