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

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

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Rewrite using trig identities

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (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

- (- a) = a

= (- 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 ( - 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, - (- 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

= - \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, - (- 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

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 = sin (x)

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

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

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

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

sec(θ)=tan(θ)

sec (θ) = tan (θ)

-sqrt(3)csc(θ)=2

- 3 csc (θ) = 2

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

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

-2+2cos(x)+4cos^2(x)=0

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

sin(A)=(6.347)/(12.7)

sin (A) = \frac{6.347}{12.7}

Frequently Asked Questions (FAQ)

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