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Popular >

cos(2x)=5-6cos^2(x)

Solution

cos (2 x) = 5 - 6 cos^{2} (x)

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

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

Solution steps

cos (2 x) = 5 - 6 cos^{2} (x)

Subtract

5 - 6 cos^{2} (x)

from both sides

cos (2 x) - 5 + 6 cos^{2} (x) = 0

Rewrite using trig identities

- 5 + cos (2 x) + 6 cos^{2} (x)

Use the Double Angle identity:

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

= - 5 + 2 cos^{2} (x) - 1 + 6 cos^{2} (x)

Simplify

- 5 + 2 cos^{2} (x) - 1 + 6 cos^{2} (x) : 8 cos^{2} (x) - 6

- 5 + 2 cos^{2} (x) - 1 + 6 cos^{2} (x)

Group like terms

= 2 cos^{2} (x) + 6 cos^{2} (x) - 5 - 1

Add similar elements:

2 cos^{2} (x) + 6 cos^{2} (x) = 8 cos^{2} (x)

= 8 cos^{2} (x) - 5 - 1

Subtract the numbers:

- 5 - 1 = - 6

= 8 cos^{2} (x) - 6

= 8 cos^{2} (x) - 6

- 6 + 8 cos^{2} (x) = 0

Solve by substitution

- 6 + 8 cos^{2} (x) = 0

Let:

cos (x) = u

- 6 + 8 u^{2} = 0

- 6 + 8 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

- 6 + 8 u^{2} = 0

Move

6

to the right side

- 6 + 8 u^{2} = 0

Add

6

to both sides

- 6 + 8 u^{2} + 6 = 0 + 6

Simplify

8 u^{2} = 6

8 u^{2} = 6

Divide both sides by

8

8 u^{2} = 6

Divide both sides by

8

\frac{8 u ^{2}}{8} = \frac{6}{8}

Simplify

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

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:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

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:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

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

Substitute back

u = cos (x)

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

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

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

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

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