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

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

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Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 3 + 2 = - 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- 1 + u - 2 u^{2} = 0

- 1 + u - 2 u^{2} = 0 : u = \frac{1}{4} - i \frac{7}{4}, u = \frac{1}{4} + i \frac{7}{4}

- 1 + u - 2 u^{2} = 0

Write in the standard form

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

- 2 u^{2} + u - 1 = 0

Solve with the quadratic formula

- 2 u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = - 1

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

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

Simplify

1^{2} - 4 (- 2) (- 1) : 7 i

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) (- 1)

Apply rule

- (- a) = a

= 1 - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 - 8

Subtract the numbers:

1 - 8 = - 7

= - 7

Apply radical rule:

- a = - 1 a

- 7 = - 1 7

= - 1 7

Apply imaginary number rule:

- 1 = i

= 7 i

u_{1, 2} = \frac{- 1 \pm 7 i}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 7 i}{2 ( - 2 )}, u_{2} = \frac{- 1 - 7 i}{2 ( - 2 )}

u = \frac{- 1 + 7 i}{2 ( - 2 )} : \frac{1}{4} - i \frac{7}{4}

\frac{- 1 + 7 i}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 7 i}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 7 i}{- 4}

Apply the fraction rule:

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

= - \frac{- 1 + 7 i}{4}

Rewrite

- \frac{- 1 + 7 i}{4}

in standard complex form:

\frac{1}{4} - \frac{7}{4} i

- \frac{- 1 + 7 i}{4}

Apply the fraction rule:

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

\frac{- 1 + 7 i}{4} = - (- \frac{1}{4}) - (\frac{7 i}{4})

= - (- \frac{1}{4}) - (\frac{7 i}{4})

Remove parentheses:

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

= \frac{1}{4} - \frac{7 i}{4}

= \frac{1}{4} - \frac{7}{4} i

u = \frac{- 1 - 7 i}{2 ( - 2 )} : \frac{1}{4} + i \frac{7}{4}

\frac{- 1 - 7 i}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 7 i}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 7 i}{- 4}

Apply the fraction rule:

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

= - \frac{- 1 - 7 i}{4}

Rewrite

- \frac{- 1 - 7 i}{4}

in standard complex form:

\frac{1}{4} + \frac{7}{4} i

- \frac{- 1 - 7 i}{4}

Apply the fraction rule:

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

\frac{- 1 - 7 i}{4} = - (- \frac{1}{4}) - (- \frac{7 i}{4})

= - (- \frac{1}{4}) - (- \frac{7 i}{4})

Apply rule

- (- a) = a

= \frac{1}{4} + \frac{7 i}{4}

= \frac{1}{4} + \frac{7}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} - i \frac{7}{4}, u = \frac{1}{4} + i \frac{7}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1}{4} - i \frac{7}{4}, cos (x) = \frac{1}{4} + i \frac{7}{4}

cos (x) = \frac{1}{4} - i \frac{7}{4}, cos (x) = \frac{1}{4} + i \frac{7}{4}

cos (x) = \frac{1}{4} - i \frac{7}{4} :

No Solution

cos (x) = \frac{1}{4} - i \frac{7}{4}

No Solution

cos (x) = \frac{1}{4} + i \frac{7}{4} :

No Solution

cos (x) = \frac{1}{4} + i \frac{7}{4}

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