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

The general solution for -sin(x)+ln(2)*cos^2(x)=0 is x=0.53713…+2pin,x=pi-0.53713…+2pin

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-sin(x)+ln(2)*cos^2(x)=0

Solution

- sin (x) + ln (2) \cdot cos^{2} (x) = 0

Solution

x = 0.53713 \dots + 2 πn, x = π - 0.53713 \dots + 2 πn

Degrees

x = 30.77540 \dots^{\circ} + 36 0^{\circ} n, x = 149.22459 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- u + (1 - u^{2}) ln (2) = 0 : u = - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}, u = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

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

Expand

- u + (1 - u^{2}) ln (2) : - u + ln (2) - ln (2) u^{2}

- u + (1 - u^{2}) ln (2)

= - u + ln (2) (1 - u^{2})

Expand

ln (2) (1 - u^{2}) : ln (2) - ln (2) u^{2}

ln (2) (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = ln (2), b = 1, c = u^{2}

= ln (2) \cdot 1 - ln (2) u^{2}

= 1 \cdot ln (2) - ln (2) u^{2}

Multiply:

1 \cdot ln (2) = ln (2)

= ln (2) - ln (2) u^{2}

= - u + ln (2) - ln (2) u^{2}

- u + ln (2) - ln (2) u^{2} = 0

Write in the standard form

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

- ln (2) u^{2} - u + ln (2) = 0

Solve with the quadratic formula

- ln (2) u^{2} - u + ln (2) = 0

Quadratic Equation Formula:

For

a = - ln (2), b = - 1, c = ln (2)

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

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

(- 1)^{2} - 4 (- ln (2)) ln (2) = 1 + 4 ln^{2} (2)

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 ln (2) ln (2)

(- 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 ln (2) ln (2) = 4 ln^{2} (2)

4 ln (2) ln (2)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

ln (2) ln (2) = ln^{1 + 1} (2)

= 4 ln^{1 + 1} (2)

Add the numbers:

1 + 1 = 2

= 4 ln^{2} (2)

= 1 + 4 ln^{2} (2)

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 1 + 4 ln ^{2} ( 2 )}{2 ( - ln ( 2 ) )} : - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}

\frac{- ( - 1 ) + 1 + 4 ln ^{2} ( 2 )}{2 ( - ln ( 2 ) )}

Remove parentheses:

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

= \frac{1 + 1 + 4 ln ^{2} ( 2 )}{- 2 ln ( 2 )}

Apply the fraction rule:

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

= - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}

u = \frac{- ( - 1 ) - 1 + 4 ln ^{2} ( 2 )}{2 ( - ln ( 2 ) )} : \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

\frac{- ( - 1 ) - 1 + 4 ln ^{2} ( 2 )}{2 ( - ln ( 2 ) )}

Remove parentheses:

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

= \frac{1 - 1 + 4 ln ^{2} ( 2 )}{- 2 ln ( 2 )}

Apply the fraction rule:

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

1 - 1 + 4 ln^{2} (2) = - (4 ln^{2} (2) + 1 - 1)

= \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

The solutions to the quadratic equation are:

u = - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}, u = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}, sin (x) = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

sin (x) = - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}, sin (x) = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

sin (x) = - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )} :

No Solution

sin (x) = - \frac{1 + 1 + 4 ln ^{2} ( 2 )}{2 ln ( 2 )}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} : x = arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn, x = π - arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn

sin (x) = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

Apply trig inverse properties

sin (x) = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

General solutions for

sin (x) = \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )}

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

x = arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn, x = π - arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn

x = arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn, x = π - arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn

Combine all the solutions

x = arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn, x = π - arcsin \frac{4 ln ^{2} ( 2 ) + 1 - 1}{2 ln ( 2 )} + 2 πn

Show solutions in decimal form

x = 0.53713 \dots + 2 πn, x = π - 0.53713 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -sin(x)+ln(2)*cos^2(x)=0 ?
The general solution for -sin(x)+ln(2)*cos^2(x)=0 is x=0.53713…+2pin,x=pi-0.53713…+2pin