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

The general solution for -e^{-sin(x)}(sin(x)+cos^2(x))=0 is x=-0.66623…+2pin,x=pi+0.66623…+2pin

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Popular >

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

Solution

- e^{- s i n (x)} (sin (x) + cos^{2} (x)) = 0

Solution

x = - 0.66623 \dots + 2 πn, x = π + 0.66623 \dots + 2 πn

Degrees

x = - 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 218.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- e^{- s i n (x)} (sin (x) + cos^{2} (x)) = 0

Rewrite using trig identities

- (cos^{2} (x) + sin (x)) e^{- s i n (x)}

Use the Pythagorean identity:

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

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

= - (1 - sin^{2} (x) + sin (x)) e^{- s i n (x)}

- (1 + sin (x) - sin^{2} (x)) e^{- s i n (x)} = 0

Solve by substitution

- (1 + sin (x) - sin^{2} (x)) e^{- s i n (x)} = 0

Let:

sin (x) = u

- (1 + u - u^{2}) e^{- u} = 0

- (1 + u - u^{2}) e^{- u} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

- (1 + u - u^{2}) e^{- u} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Solve

e^{- u} = 0 :

No Solution for

u \in R

e^{- u} = 0

a^{f (u)}

cannot be zero or negative for

u \in R

No Solution for u \in R

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}

sin (x) = - \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}

sin (x) = - \frac{- 1 + 5}{2} : x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = - \frac{- 1 + 5}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 + 5}{2}

General solutions for

sin (x) = - \frac{- 1 + 5}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{1 + 5}{2} :

No Solution

sin (x) = \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = - 0.66623 \dots + 2 πn, x = π + 0.66623 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -e^{-sin(x)}(sin(x)+cos^2(x))=0 ?
The general solution for -e^{-sin(x)}(sin(x)+cos^2(x))=0 is x=-0.66623…+2pin,x=pi+0.66623…+2pin