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

The general solution for cos^2(x)-sin^2(x)=0 is x= pi/4+pin,x=(3pi)/4+pin

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

Solution

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

Solution

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

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

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) = 0

General solutions for

cos (2 x) = 0

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

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

Solve

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

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

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

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

Popular Examples

tan(θ)=2sin(θ)

tan (θ) = 2 sin (θ)

cos(x)=(-1)/2

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

csc(x)-2=0

csc (x) - 2 = 0

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

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

tan(x)=2sin(x)

tan (x) = 2 sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-sin^2(x)=0 ?
The general solution for cos^2(x)-sin^2(x)=0 is x= pi/4+pin,x=(3pi)/4+pin