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cos^2(x)+cos^2(2x)=1

Solution

cos^{2} (x) + cos^{2} (2 x) = 1

Solution

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) + cos^{2} (2 x) = 1

Subtract

1

from both sides

cos^{2} (x) + cos^{2} (2 x) - 1 = 0

Rewrite using trig identities

- 1 + cos^{2} (2 x) + cos^{2} (x)

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= - 1 + (2 cos^{2} (x) - 1)^{2} + cos^{2} (x)

Simplify

- 1 + (2 cos^{2} (x) - 1)^{2} + cos^{2} (x) : 4 cos^{4} (x) - 3 cos^{2} (x)

- 1 + (2 cos^{2} (x) - 1)^{2} + cos^{2} (x)

(2 cos^{2} (x) - 1)^{2} : 4 cos^{4} (x) - 4 cos^{2} (x) + 1

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 2 cos^{2} (x), b = 1

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

Simplify

(2 cos^{2} (x))^{2} - 2 \cdot 2 cos^{2} (x) \cdot 1 + 1^{2} : 4 cos^{4} (x) - 4 cos^{2} (x) + 1

(2 cos^{2} (x))^{2} - 2 \cdot 2 cos^{2} (x) \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

(2 cos^{2} (x))^{2} = 4 cos^{4} (x)

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(cos^{2} (x))^{2} : cos^{4} (x)

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= cos^{2 \cdot 2} (x)

Multiply the numbers:

2 \cdot 2 = 4

= cos^{4} (x)

= 2^{2} cos^{4} (x)

2^{2} = 4

= 4 cos^{4} (x)

2 \cdot 2 cos^{2} (x) \cdot 1 = 4 cos^{2} (x)

2 \cdot 2 cos^{2} (x) \cdot 1

Multiply the numbers:

2 \cdot 2 \cdot 1 = 4

= 4 cos^{2} (x)

= 4 cos^{4} (x) - 4 cos^{2} (x) + 1

= 4 cos^{4} (x) - 4 cos^{2} (x) + 1

= - 1 + 4 cos^{4} (x) - 4 cos^{2} (x) + 1 + cos^{2} (x)

Simplify

- 1 + 4 cos^{4} (x) - 4 cos^{2} (x) + 1 + cos^{2} (x) : 4 cos^{4} (x) - 3 cos^{2} (x)

- 1 + 4 cos^{4} (x) - 4 cos^{2} (x) + 1 + cos^{2} (x)

Group like terms

= 4 cos^{4} (x) - 4 cos^{2} (x) + cos^{2} (x) - 1 + 1

Add similar elements:

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

= 4 cos^{4} (x) - 3 cos^{2} (x) - 1 + 1

- 1 + 1 = 0

= 4 cos^{4} (x) - 3 cos^{2} (x)

= 4 cos^{4} (x) - 3 cos^{2} (x)

= 4 cos^{4} (x) - 3 cos^{2} (x)

- 3 cos^{2} (x) + 4 cos^{4} (x) = 0

Solve by substitution

- 3 cos^{2} (x) + 4 cos^{4} (x) = 0

Let:

cos (x) = u

- 3 u^{2} + 4 u^{4} = 0

- 3 u^{2} + 4 u^{4} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}, u = 0

- 3 u^{2} + 4 u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 u^{4} - 3 u^{2} = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

4 v^{2} - 3 v = 0

Solve

4 v^{2} - 3 v = 0 : v = \frac{3}{4}, v = 0

4 v^{2} - 3 v = 0

Solve with the quadratic formula

4 v^{2} - 3 v = 0

Quadratic Equation Formula:

For

a = 4, b = - 3, c = 0

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 4 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 4 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 3

v_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 4}, v_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 4}

v = \frac{- ( - 3 ) + 3}{2 \cdot 4} : \frac{3}{4}

\frac{- ( - 3 ) + 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 + 3}{2 \cdot 4}

Add the numbers:

3 + 3 = 6

= \frac{6}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

v = \frac{- ( - 3 ) - 3}{2 \cdot 4} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 - 3}{2 \cdot 4}

Subtract the numbers:

3 - 3 = 0

= \frac{0}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{0}{8}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

v = \frac{3}{4}, v = 0

v = \frac{3}{4}, v = 0

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{3}{4} : u = \frac{3}{2}, u = - \frac{3}{2}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{3}{4}, u = - \frac{3}{4}

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

\frac{3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

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

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The solutions are

u = \frac{3}{2}, u = - \frac{3}{2}, u = 0

Substitute back

u = cos (x)

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

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

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

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

General solutions for

cos (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}

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

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

cos (x) = - \frac{3}{2} : x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

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

General solutions for

cos (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}

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

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

cos (x) = 0

General solutions for

cos (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}

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

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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Graph

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