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

Apply minus-plus rules

- (- a) = a

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

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

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

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

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

Simplify

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

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 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 \cdot 2}

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

= \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}

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

= \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}

= \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

Popular Examples

2tan^2(2x)-tan(2x)=3

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

sin(x)= 15/31

sin (x) = \frac{15}{31}

4cot(θ)-1=0

4 cot (θ) - 1 = 0

solvefor x,4cos^2(x)=5-4sin(x)solve for

x, 4 cos^{2} (x) = 5 - 4 sin (x)

cos(x)= 2/3 ,-(5pi)/2 <x<2pi

cos (x) = \frac{2}{3}, - \frac{5 π}{2} < x < 2 π

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}