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

The general solution for cos^2(x)-sin^2(x)=(sqrt(3))/2 is x= pi/(12)+pin,x=(11pi)/(12)+pin

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cos^2(x)-sin^2(x)=(sqrt(3))/2

Solution

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

Solution

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

Degrees

x = 1 5^{\circ} + 18 0^{\circ} n, x = 16 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= cos (2 x)

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

General solutions for

cos (2 x) = \frac{3}{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 x = \frac{π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

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

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{12} + πn

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

Popular Examples

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2 sin^{2} (x) + 1 = 4 sin (x)

1/4 =cos(x)

\frac{1}{4} = cos (x)

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

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

4tan^2(θ)+16tan(θ)=7tan(θ)-2

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

10(1-cos(θ))=sin^2(θ)

10 (1 - cos (θ)) = sin^{2} (θ)

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-sin^2(x)=(sqrt(3))/2 ?
The general solution for cos^2(x)-sin^2(x)=(sqrt(3))/2 is x= pi/(12)+pin,x=(11pi)/(12)+pin