What is the general solution for cos^2(2x)-3sin^2(x)-1=0 ?

The general solution for cos^2(2x)-3sin^2(x)-1=0 is x=2pin,x=pi+2pin

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cos^2(2x)-3sin^2(x)-1=0

Solution

cos^{2} (2 x) - 3 sin^{2} (x) - 1 = 0

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (2 x) - 3 sin^{2} (x) - 1 = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

- 1 + (1 - 2 sin^{2} (x))^{2} - 3 sin^{2} (x) : 4 sin^{4} (x) - 7 sin^{2} (x)

- 1 + (1 - 2 sin^{2} (x))^{2} - 3 sin^{2} (x)

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply the numbers:

2 \cdot 1 \cdot 2 = 4

= 4 sin^{2} (x)

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

(2 sin^{2} (x))^{2}

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (x)

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

2^{2} = 4

= 4 sin^{4} (x)

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

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

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

Simplify

- 1 + 1 - 4 sin^{2} (x) + 4 sin^{4} (x) - 3 sin^{2} (x) : 4 sin^{4} (x) - 7 sin^{2} (x)

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

Group like terms

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

Add similar elements:

- 4 sin^{2} (x) - 3 sin^{2} (x) = - 7 sin^{2} (x)

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

- 1 + 1 = 0

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

4 u^{4} - 7 u^{2} = 0

4 u^{4} - 7 u^{2} = 0 : u = \frac{7}{2}, u = - \frac{7}{2}, u = 0

4 u^{4} - 7 u^{2} = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

4 v^{2} - 7 v = 0

Solve

4 v^{2} - 7 v = 0 : v = \frac{7}{4}, v = 0

4 v^{2} - 7 v = 0

Solve with the quadratic formula

4 v^{2} - 7 v = 0

Quadratic Equation Formula:

For

a = 4, b = - 7, c = 0

v_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 7)^{2} - 4 \cdot 4 \cdot 0 = 7

(- 7)^{2} - 4 \cdot 4 \cdot 0

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 4 \cdot 0

Apply rule

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

Apply radical rule:

assuming

a \geq 0

= 7

v_{1, 2} = \frac{- ( - 7 ) \pm 7}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- ( - 7 ) + 7}{2 \cdot 4}, v_{2} = \frac{- ( - 7 ) - 7}{2 \cdot 4}

v = \frac{- ( - 7 ) + 7}{2 \cdot 4} : \frac{7}{4}

\frac{- ( - 7 ) + 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{7 + 7}{2 \cdot 4}

Add the numbers:

7 + 7 = 14

= \frac{14}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{14}{8}

Cancel the common factor:

2

= \frac{7}{4}

v = \frac{- ( - 7 ) - 7}{2 \cdot 4} : 0

\frac{- ( - 7 ) - 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{7 - 7}{2 \cdot 4}

Subtract the numbers:

7 - 7 = 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{7}{4}, v = 0

v = \frac{7}{4}, v = 0

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{7}{4} : u = \frac{7}{2}, u = - \frac{7}{2}

u^{2} = \frac{7}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{7}{4}, u = - \frac{7}{4}

\frac{7}{4} = \frac{7}{2}

\frac{7}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7}{2}

- \frac{7}{4} = - \frac{7}{2}

- \frac{7}{4}

Simplify

\frac{7}{4} : \frac{7}{2}

\frac{7}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7}{2}

= - \frac{7}{2}

u = \frac{7}{2}, u = - \frac{7}{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{7}{2}, u = - \frac{7}{2}, u = 0

Substitute back

u = sin (x)

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}, sin (x) = 0

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}, sin (x) = 0

sin (x) = \frac{7}{2} :

No Solution

sin (x) = \frac{7}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{7}{2} :

No Solution

sin (x) = - \frac{7}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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sin (- t) = \frac{3}{10}

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(2x)-3sin^2(x)-1=0 ?
The general solution for cos^2(2x)-3sin^2(x)-1=0 is x=2pin,x=pi+2pin