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Popular >

4cos(2t)=4sin^2(t)+1

Solution

4 cos (2 t) = 4 sin^{2} (t) + 1

Solution

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

Degrees

t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n, t = 21 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (2 t) = 4 sin^{2} (t) + 1

Subtract

4 sin^{2} (t) + 1

from both sides

4 cos (2 t) - 4 sin^{2} (t) - 1 = 0

Rewrite using trig identities

- 1 + 4 cos (2 t) - 4 sin^{2} (t)

Use the Double Angle identity:

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

= - 1 + 4 (1 - 2 sin^{2} (t)) - 4 sin^{2} (t)

Simplify

- 1 + 4 (1 - 2 sin^{2} (t)) - 4 sin^{2} (t) : - 12 sin^{2} (t) + 3

- 1 + 4 (1 - 2 sin^{2} (t)) - 4 sin^{2} (t)

Expand

4 (1 - 2 sin^{2} (t)) : 4 - 8 sin^{2} (t)

4 (1 - 2 sin^{2} (t))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = 2 sin^{2} (t)

= 4 \cdot 1 - 4 \cdot 2 sin^{2} (t)

Simplify

4 \cdot 1 - 4 \cdot 2 sin^{2} (t) : 4 - 8 sin^{2} (t)

4 \cdot 1 - 4 \cdot 2 sin^{2} (t)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 \cdot 2 sin^{2} (t)

Multiply the numbers:

4 \cdot 2 = 8

= 4 - 8 sin^{2} (t)

= 4 - 8 sin^{2} (t)

= - 1 + 4 - 8 sin^{2} (t) - 4 sin^{2} (t)

Simplify

- 1 + 4 - 8 sin^{2} (t) - 4 sin^{2} (t) : - 12 sin^{2} (t) + 3

- 1 + 4 - 8 sin^{2} (t) - 4 sin^{2} (t)

Add similar elements:

- 8 sin^{2} (t) - 4 sin^{2} (t) = - 12 sin^{2} (t)

= - 1 + 4 - 12 sin^{2} (t)

Add/Subtract the numbers:

- 1 + 4 = 3

= - 12 sin^{2} (t) + 3

= - 12 sin^{2} (t) + 3

= - 12 sin^{2} (t) + 3

3 - 12 sin^{2} (t) = 0

Solve by substitution

3 - 12 sin^{2} (t) = 0

Let:

sin (t) = u

3 - 12 u^{2} = 0

3 - 12 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

3 - 12 u^{2} = 0

Move

3

to the right side

3 - 12 u^{2} = 0

Subtract

3

from both sides

3 - 12 u^{2} - 3 = 0 - 3

Simplify

- 12 u^{2} = - 3

- 12 u^{2} = - 3

Divide both sides by

- 12

- 12 u^{2} = - 3

Divide both sides by

- 12

\frac{- 12 u ^{2}}{- 12} = \frac{- 3}{- 12}

Simplify

\frac{- 12 u ^{2}}{- 12} = \frac{- 3}{- 12}

Simplify

\frac{- 12 u ^{2}}{- 12} : u^{2}

\frac{- 12 u ^{2}}{- 12}

Apply the fraction rule:

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

= \frac{12 u ^{2}}{12}

Divide the numbers:

\frac{12}{12} = 1

= u^{2}

Simplify

\frac{- 3}{- 12} : \frac{1}{4}

\frac{- 3}{- 12}

Apply the fraction rule:

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

= \frac{3}{12}

Cancel the common factor:

3

= \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = sin (t)

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{2}

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{2}

sin (t) = \frac{1}{2} : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

sin (t) = \frac{1}{2}

General solutions for

sin (t) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

sin (t) = - \frac{1}{2}

General solutions for

sin (t) = - \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Combine all the solutions

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

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