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Popular >

4sin^4(x)+12cos^2(x)-7=0

Solution

4 sin^{4} (x) + 12 cos^{2} (x) - 7 = 0

Solution

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{4} (x) + 12 cos^{2} (x) - 7 = 0

Rewrite using trig identities

- 7 + 12 cos^{2} (x) + 4 sin^{4} (x)

Use the Pythagorean identity:

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

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

= - 7 + 12 (1 - sin^{2} (x)) + 4 sin^{4} (x)

Simplify

- 7 + 12 (1 - sin^{2} (x)) + 4 sin^{4} (x) : 4 sin^{4} (x) - 12 sin^{2} (x) + 5

- 7 + 12 (1 - sin^{2} (x)) + 4 sin^{4} (x)

Expand

12 (1 - sin^{2} (x)) : 12 - 12 sin^{2} (x)

12 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 12, b = 1, c = sin^{2} (x)

= 12 \cdot 1 - 12 sin^{2} (x)

Multiply the numbers:

12 \cdot 1 = 12

= 12 - 12 sin^{2} (x)

= - 7 + 12 - 12 sin^{2} (x) + 4 sin^{4} (x)

Add/Subtract the numbers:

- 7 + 12 = 5

= 4 sin^{4} (x) - 12 sin^{2} (x) + 5

= 4 sin^{4} (x) - 12 sin^{2} (x) + 5

5 - 12 sin^{2} (x) + 4 sin^{4} (x) = 0

Solve by substitution

5 - 12 sin^{2} (x) + 4 sin^{4} (x) = 0

Let:

sin (x) = u

5 - 12 u^{2} + 4 u^{4} = 0

5 - 12 u^{2} + 4 u^{4} = 0 : u = \frac{5}{2}, u = - \frac{5}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

5 - 12 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} - 12 u^{2} + 5 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

4 v^{2} - 12 v + 5 = 0

Solve

4 v^{2} - 12 v + 5 = 0 : v = \frac{5}{2}, v = \frac{1}{2}

4 v^{2} - 12 v + 5 = 0

Solve with the quadratic formula

4 v^{2} - 12 v + 5 = 0

Quadratic Equation Formula:

For

a = 4, b = - 12, c = 5

v_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 \cdot 5 = 8

(- 12)^{2} - 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 1 2^{2} - 80

1 2^{2} = 144

= 144 - 80

Subtract the numbers:

144 - 80 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 12 ) \pm 8}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- ( - 12 ) + 8}{2 \cdot 4}, v_{2} = \frac{- ( - 12 ) - 8}{2 \cdot 4}

v = \frac{- ( - 12 ) + 8}{2 \cdot 4} : \frac{5}{2}

\frac{- ( - 12 ) + 8}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{12 + 8}{2 \cdot 4}

Add the numbers:

12 + 8 = 20

= \frac{20}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{20}{8}

Cancel the common factor:

4

= \frac{5}{2}

v = \frac{- ( - 12 ) - 8}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 12 ) - 8}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{12 - 8}{2 \cdot 4}

Subtract the numbers:

12 - 8 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

v = \frac{5}{2}, v = \frac{1}{2}

v = \frac{5}{2}, v = \frac{1}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{5}{2} : u = \frac{5}{2}, u = - \frac{5}{2}

u^{2} = \frac{5}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{5}{2}, u = - \frac{5}{2}

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

The solutions are

u = \frac{5}{2}, u = - \frac{5}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = \frac{5}{2}, sin (x) = - \frac{5}{2}, sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{5}{2}, sin (x) = - \frac{5}{2}, sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{5}{2} :

No Solution

sin (x) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{5}{2} :

No Solution

sin (x) = - \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

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

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

General solutions for

sin (x) = \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}

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

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

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

General solutions for

sin (x) = - \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}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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