What is the general solution for tan(2x)(16cos^4(x)-16cos^2(x)+3)=0 ?

The general solution for tan(2x)(16cos^4(x)-16cos^2(x)+3)=0 is

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tan(2x)(16cos^4(x)-16cos^2(x)+3)=0

Solution

tan (2 x) (16 cos^{4} (x) - 16 cos^{2} (x) + 3) = 0

Solution

x = \frac{πn}{2}, 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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 0^{\circ} + 9 0^{\circ} n, 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 = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (2 x) (16 cos^{4} (x) - 16 cos^{2} (x) + 3) = 0

Solving each part separately

tan (2 x) = 0 or 16 cos^{4} (x) - 16 cos^{2} (x) + 3 = 0

tan (2 x) = 0 : x = \frac{πn}{2}

tan (2 x) = 0

General solutions for

tan (2 x) = 0

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x = 0 + πn

2 x = 0 + πn

Solve

2 x = 0 + πn : x = \frac{πn}{2}

2 x = 0 + πn

0 + πn = πn

2 x = πn

Divide both sides by

2

2 x = πn

Divide both sides by

2

\frac{2 x}{2} = \frac{πn}{2}

Simplify

x = \frac{πn}{2}

x = \frac{πn}{2}

x = \frac{πn}{2}

16 cos^{4} (x) - 16 cos^{2} (x) + 3 = 0 : 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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

16 cos^{4} (x) - 16 cos^{2} (x) + 3 = 0

Solve by substitution

16 cos^{4} (x) - 16 cos^{2} (x) + 3 = 0

Let:

cos (x) = u

16 u^{4} - 16 u^{2} + 3 = 0

16 u^{4} - 16 u^{2} + 3 = 0 : u = \frac{3}{2}, u = - \frac{3}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

16 u^{4} - 16 u^{2} + 3 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

16 v^{2} - 16 v + 3 = 0

Solve

16 v^{2} - 16 v + 3 = 0 : v = \frac{3}{4}, v = \frac{1}{4}

16 v^{2} - 16 v + 3 = 0

Solve with the quadratic formula

16 v^{2} - 16 v + 3 = 0

Quadratic Equation Formula:

For

a = 16, b = - 16, c = 3

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 16 \cdot 3}{2 \cdot 16}

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 16 \cdot 3}{2 \cdot 16}

(- 16)^{2} - 4 \cdot 16 \cdot 3 = 8

(- 16)^{2} - 4 \cdot 16 \cdot 3

Apply exponent rule:

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

n

is even

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 16 \cdot 3

Multiply the numbers:

4 \cdot 16 \cdot 3 = 192

= 1 6^{2} - 192

1 6^{2} = 256

= 256 - 192

Subtract the numbers:

256 - 192 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 16 ) \pm 8}{2 \cdot 16}

Separate the solutions

v_{1} = \frac{- ( - 16 ) + 8}{2 \cdot 16}, v_{2} = \frac{- ( - 16 ) - 8}{2 \cdot 16}

v = \frac{- ( - 16 ) + 8}{2 \cdot 16} : \frac{3}{4}

\frac{- ( - 16 ) + 8}{2 \cdot 16}

Apply rule

- (- a) = a

= \frac{16 + 8}{2 \cdot 16}

Add the numbers:

16 + 8 = 24

= \frac{24}{2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{24}{32}

Cancel the common factor:

8

= \frac{3}{4}

v = \frac{- ( - 16 ) - 8}{2 \cdot 16} : \frac{1}{4}

\frac{- ( - 16 ) - 8}{2 \cdot 16}

Apply rule

- (- a) = a

= \frac{16 - 8}{2 \cdot 16}

Subtract the numbers:

16 - 8 = 8

= \frac{8}{2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{8}{32}

Cancel the common factor:

8

= \frac{1}{4}

The solutions to the quadratic equation are:

v = \frac{3}{4}, v = \frac{1}{4}

v = \frac{3}{4}, v = \frac{1}{4}

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:

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}

Solve

u^{2} = \frac{1}{4} : u = \frac{1}{2}, u = - \frac{1}{2}

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}

The solutions are

u = \frac{3}{2}, u = - \frac{3}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}, cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}, cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

cos (x) = \frac{1}{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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

cos (x) = - \frac{1}{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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = \frac{πn}{2}, 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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)(16cos^4(x)-16cos^2(x)+3)=0 ?
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