What is the general solution for sin^2(2x)-cos^2(2x)= 1/2 ?

The general solution for sin^2(2x)-cos^2(2x)= 1/2 is x= pi/6+(pin)/2 ,x= pi/3+(pin)/2

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sin^2(2x)-cos^2(2x)= 1/2

Solution

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

Solution

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

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

= - cos (2 \cdot 2 x)

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

Divide both sides by

- 1

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

Divide both sides by

- 1

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

Simplify

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

Simplify

\frac{- cos ( 2 \cdot 2 x )}{- 1} : cos (2 \cdot 2 x)

\frac{- cos ( 2 \cdot 2 x )}{- 1}

Apply the fraction rule:

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

= \frac{cos ( 2 \cdot 2 x )}{1}

Apply rule

\frac{a}{1} = a

= cos (2 \cdot 2 x)

Simplify

\frac{\frac{1}{2}}{- 1} : - \frac{1}{2}

\frac{\frac{1}{2}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{1}{2}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{1}{2}}{1} = \frac{1}{2}

= - \frac{1}{2}

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

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

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

General solutions for

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

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 \cdot 2 x = \frac{2 π}{3} + 2 πn, 2 \cdot 2 x = \frac{4 π}{3} + 2 πn

2 \cdot 2 x = \frac{2 π}{3} + 2 πn, 2 \cdot 2 x = \frac{4 π}{3} + 2 πn

Solve

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

2 \cdot 2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

4

2 \cdot 2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

4

\frac{2 \cdot 2 x}{4} = \frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{2 \cdot 2 x}{4} = \frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{6} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{2 π}{12}

Cancel the common factor:

2

= \frac{π}{6}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \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}

Solve

2 \cdot 2 x = \frac{4 π}{3} + 2 πn : x = \frac{π}{3} + \frac{πn}{2}

2 \cdot 2 x = \frac{4 π}{3} + 2 πn

Divide both sides by

4

2 \cdot 2 x = \frac{4 π}{3} + 2 πn

Divide both sides by

4

\frac{2 \cdot 2 x}{4} = \frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{2 \cdot 2 x}{4} = \frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{3} + \frac{πn}{2}

\frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{4 π}{3}}{4} = \frac{π}{3}

\frac{\frac{4 π}{3}}{4}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{4 π}{12}

Cancel the common factor:

4

= \frac{π}{3}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

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

Popular Examples

sin(x)=(1.2)/(sqrt(10))

sin (x) = \frac{1.2}{10}

solvefor t,x=-3cos(pit)solve for

t, x = - 3 cos (π t)

(-1)/2 =cos(x)

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

cos(2t)-sin(t)=0.5,0<= t<= 2pi

cos (2 t) - sin (t) = 0.5, 0 \leq t \leq 2 π

sin(2x)=0.1

sin (2 x) = 0.1

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(2x)-cos^2(2x)= 1/2 ?
The general solution for sin^2(2x)-cos^2(2x)= 1/2 is x= pi/6+(pin)/2 ,x= pi/3+(pin)/2