What is the general solution for (sin(x)+sin^2(x))/2 =0.5 ?

The general solution for (sin(x)+sin^2(x))/2 =0.5 is x=0.66623…+2pin,x=pi-0.66623…+2pin

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(sin(x)+sin^2(x))/2 =0.5

Solution

\frac{sin ( x ) + sin ^{2} ( x )}{2} = 0.5

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 = 141.82729 \dots^{\circ} + 36 0^{\circ} n

Solution steps

\frac{sin ( x ) + sin ^{2} ( x )}{2} = 0.5

Solve by substitution

\frac{sin ( x ) + sin ^{2} ( x )}{2} = 0.5

Let:

sin (x) = u

\frac{u + u ^{2}}{2} = 0.5

\frac{u + u ^{2}}{2} = 0.5 : u = \frac{- 1 + 5}{2}, u = - \frac{1 + 5}{2}

\frac{u + u ^{2}}{2} = 0.5

Multiply both sides by

10

\frac{u + u ^{2}}{2} = 0.5

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

\frac{u + u ^{2}}{2} \cdot 10 = 0.5 \cdot 10

Refine

5 (u^{2} + u) = 5

5 (u^{2} + u) = 5

Expand

5 (u^{2} + u) : 5 u^{2} + 5 u

5 (u^{2} + u)

Apply the distributive law:

a (b + c) = ab + a c

a = 5, b = u^{2}, c = u

= 5 u^{2} + 5 u

5 u^{2} + 5 u = 5

Move

5

to the left side

5 u^{2} + 5 u = 5

Subtract

5

from both sides

5 u^{2} + 5 u - 5 = 5 - 5

Simplify

5 u^{2} + 5 u - 5 = 0

5 u^{2} + 5 u - 5 = 0

Solve with the quadratic formula

5 u^{2} + 5 u - 5 = 0

Quadratic Equation Formula:

For

a = 5, b = 5, c = - 5

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

5^{2} - 4 \cdot 5 (- 5) = 55

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 5 \cdot 5

Multiply the numbers:

4 \cdot 5 \cdot 5 = 100

= 5^{2} + 100

5^{2} = 25

= 25 + 100

Add the numbers:

25 + 100 = 125

= 125

Prime factorization of

125 : 5^{3}

125

125

divides by

5125 = 25 \cdot 5

= 5 \cdot 25

25

divides by

525 = 5 \cdot 5

= 5 \cdot 5 \cdot 5

5

is a prime number, therefore no further factorization is possible

= 5 \cdot 5 \cdot 5

= 5^{3}

= 5^{3}

Apply exponent rule:

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

= 5^{2} \cdot 5

Apply radical rule:

= 5 5^{2}

Apply radical rule:

5^{2} = 5

= 55

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 5 + 5 5}{10}

Factor

- 5 + 55 : 5 (- 1 + 5)

- 5 + 55

Rewrite as

= - 5 \cdot 1 + 55

Factor out common term

5

= 5 (- 1 + 5)

= \frac{5 ( - 1 + 5 )}{10}

Cancel the common factor:

5

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

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

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 5 - 5 5}{10}

Factor

- 5 - 55 : - 5 (1 + 5)

- 5 - 55

Rewrite as

= - 5 \cdot 1 - 55

Factor out common term

5

= - 5 (1 + 5)

= - \frac{5 ( 1 + 5 )}{10}

Cancel the common factor:

5

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

The solutions to the quadratic equation are:

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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arctan (1 - x) + arctan (1 + x) = arctan (\frac{1}{8})

5sin(4x)=2

5 sin (4 x) = 2

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solvefor x,log_{10}(y)=arctan(x)+c solve for

x, lo g_{10} (y) = arctan (x) + c

Frequently Asked Questions (FAQ)

What is the general solution for (sin(x)+sin^2(x))/2 =0.5 ?
The general solution for (sin(x)+sin^2(x))/2 =0.5 is x=0.66623…+2pin,x=pi-0.66623…+2pin