What is the general solution for (sin(x))^2+(1-sqrt(2))sin(x)-sqrt(2)=0 ?

The general solution for (sin(x))^2+(1-sqrt(2))sin(x)-sqrt(2)=0 is x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(sin(x))^2+(1-sqrt(2))sin(x)-sqrt(2)=0

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 1 - 2, c = - 2

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

= (1 - 2)^{2} + 42

Expand

(1 - 2)^{2} + 42 : 3 + 22

(1 - 2)^{2} + 42

(1 - 2)^{2} : 3 - 22

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = 2

= 1^{2} - 2 \cdot 1 \cdot 2 + (2)^{2}

Simplify

1^{2} - 2 \cdot 1 \cdot 2 + (2)^{2} : 3 - 22

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot 2 + (2)^{2}

2 \cdot 1 \cdot 2 = 22

2 \cdot 1 \cdot 2

Multiply the numbers:

2 \cdot 1 = 2

= 22

(2)^{2} = 2

(2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= 2

= 1 - 22 + 2

Add the numbers:

1 + 2 = 3

= 3 - 22

= 3 - 22

= 3 - 22 + 42

Add similar elements:

- 22 + 42 = 22

= 3 + 22

= 3 + 22

= 2 + 22 + 1

= (2)^{2} + 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} + 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} + 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(2)^{2} + 22 \cdot 1 + 1^{2} = (2 + 1)^{2}

= (2 + 1)^{2}

Apply radical rule:

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

= 2 + 1

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Expand

- (1 - 2) + 2 + 1 : 22

- (1 - 2) + 2 + 1

- (1 - 2) : - 1 + 2

- (1 - 2)

Distribute parentheses

= - (1) - (- 2)

Apply minus-plus rules

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

= - 1 + 2

= - 1 + 2 + 2 + 1

Simplify

- 1 + 2 + 2 + 1 : 22

- 1 + 2 + 2 + 1

Add similar elements:

2 + 2 = 22

= - 1 + 22 + 1

- 1 + 1 = 0

= 22

= 22

= \frac{2 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Expand

- (1 - 2) - (2 + 1) : - 2

- (1 - 2) - (2 + 1)

- (1 - 2) : - 1 + 2

- (1 - 2)

Distribute parentheses

= - (1) - (- 2)

Apply minus-plus rules

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

= - 1 + 2

= - 1 + 2 - (2 + 1)

- (2 + 1) : - 2 - 1

- (2 + 1)

Distribute parentheses

= - (2) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 2 - 1

= - 1 + 2 - 2 - 1

Simplify

- 1 + 2 - 2 - 1 : - 2

- 1 + 2 - 2 - 1

Add similar elements:

2 - 2 = 0

= - 1 - 1

Subtract the numbers:

- 1 - 1 = - 2

= - 2

= - 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 2, u = - 1

Substitute back

u = sin (x)

sin (x) = 2, sin (x) = - 1

sin (x) = 2, sin (x) = - 1

sin (x) = 2 :

No Solution

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

4sin^2(A)-1=0

4 sin^{2} (A) - 1 = 0

tan(θ)= 22/7

tan (θ) = \frac{22}{7}

solvefor u,x=4tanh(u)solve for

u, x = 4 tanh (u)

tan(x)= 2/7

tan (x) = \frac{2}{7}

sin^2(θ)+cos(θ)-cos^2(θ)=0

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

Frequently Asked Questions (FAQ)

What is the general solution for (sin(x))^2+(1-sqrt(2))sin(x)-sqrt(2)=0 ?
The general solution for (sin(x))^2+(1-sqrt(2))sin(x)-sqrt(2)=0 is x=(3pi)/2+2pin