What is the general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 ?

Q: What is the general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 ?

The general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 is x=90,x=30,x=150

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(x)-3sin(x)+1=0,0<= x<= 180

Solution

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Solution

x = 9 0^{\circ}, x = 3 0^{\circ}, x = 15 0^{\circ}

Radians

x = \frac{π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Solve by substitution

2 sin^{2} (x) - 3 sin (x) + 1 = 0

Let:

sin (x) = u

2 u^{2} - 3 u + 1 = 0

2 u^{2} - 3 u + 1 = 0 : u = 1, u = \frac{1}{2}

2 u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

(- 3)^{2} - 4 \cdot 2 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 2}

u = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

\frac{- ( - 3 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

Add the numbers:

3 + 1 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{2}

Substitute back

u = sin (x)

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

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

sin (x) = 1, 0 \leq x \leq 18 0^{\circ} : x = 9 0^{\circ}

sin (x) = 1, 0 \leq x \leq 18 0^{\circ}

General solutions for

sin (x) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 9 0^{\circ} + 36 0^{\circ} n

x = 9 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = 9 0^{\circ}

sin (x) = \frac{1}{2}, 0 \leq x \leq 18 0^{\circ} : x = 3 0^{\circ}, x = 15 0^{\circ}

sin (x) = \frac{1}{2}, 0 \leq x \leq 18 0^{\circ}

General solutions for

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

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = 3 0^{\circ}, x = 15 0^{\circ}

Combine all the solutions

x = 9 0^{\circ}, x = 3 0^{\circ}, x = 15 0^{\circ}

Popular Examples

sin^2(θ)+cos(θ)=2

sin^{2} (θ) + cos (θ) = 2

cot(2x)-1=0

cot (2 x) - 1 = 0

tan(2θ)-tan(θ)=0,0<θ<2pi

tan (2 θ) - tan (θ) = 0, 0 < θ < 2 π

-8cos(x)=-6sec(x)+8

- 8 cos (x) = - 6 sec (x) + 8

cos(2θ)+sin(θ)=1

cos (2 θ) + sin (θ) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 ?
The general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 is x=90,x=30,x=150