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

The general solution for sin^2(x)-3cos(x)-4=0 is No Solution for x\in\mathbb{R}

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Popular >

sin^2(x)-3cos(x)-4=0

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - cos^{2} (x) - 3 cos (x) - 3

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = - 3, c = - 3

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 3 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 3 )}{2 ( - 1 )}

Simplify

(- 3)^{2} - 4 (- 1) (- 3) : 3 i

(- 3)^{2} - 4 (- 1) (- 3)

Apply rule

- (- a) = a

= (- 3)^{2} - 4 \cdot 1 \cdot 3

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 3^{2} - 12

Apply imaginary number rule:

- a = i a

= i 12 - 3^{2}

- 3^{2} + 12 = 3

- 3^{2} + 12

3^{2} = 9

= - 9 + 12

Add/Subtract the numbers:

- 9 + 12 = 3

= 3

= 3 i

u_{1, 2} = \frac{- ( - 3 ) \pm 3 i}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3 i}{2 ( - 1 )}, u_{2} = \frac{- ( - 3 ) - 3 i}{2 ( - 1 )}

u = \frac{- ( - 3 ) + 3 i}{2 ( - 1 )} : - \frac{3}{2} - i \frac{3}{2}

\frac{- ( - 3 ) + 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 + 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 + 3 i}{- 2}

Apply the fraction rule:

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

= - \frac{3 + 3 i}{2}

Rewrite

- \frac{3 + 3 i}{2}

in standard complex form:

- \frac{3}{2} - \frac{3}{2} i

- \frac{3 + 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{3 + 3 i}{2} = - (\frac{3}{2}) - (\frac{3 i}{2})

= - (\frac{3}{2}) - (\frac{3 i}{2})

Remove parentheses:

(a) = a

= - \frac{3}{2} - \frac{3 i}{2}

= - \frac{3}{2} - \frac{3}{2} i

u = \frac{- ( - 3 ) - 3 i}{2 ( - 1 )} : - \frac{3}{2} + i \frac{3}{2}

\frac{- ( - 3 ) - 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 - 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 - 3 i}{- 2}

Apply the fraction rule:

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

= - \frac{3 - 3 i}{2}

Rewrite

- \frac{3 - 3 i}{2}

in standard complex form:

- \frac{3}{2} + \frac{3}{2} i

- \frac{3 - 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{3 - 3 i}{2} = - (\frac{3}{2}) - (- \frac{3 i}{2})

= - (\frac{3}{2}) - (- \frac{3 i}{2})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{3}{2} + \frac{3 i}{2}

= - \frac{3}{2} + \frac{3}{2} i

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

No Solution

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

No Solution

cos (x) = - \frac{3}{2} + i \frac{3}{2} :

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-3cos(x)-4=0 ?
The general solution for sin^2(x)-3cos(x)-4=0 is No Solution for x\in\mathbb{R}