What is the general solution for sin^2(x)-7sin(x)=0 ?

The general solution for sin^2(x)-7sin(x)=0 is x=2pin,x=pi+2pin

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sin^2(x)-7sin(x)=0

Solution

sin^{2} (x) - 7 sin (x) = 0

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) - 7 sin (x) = 0

Solve by substitution

sin^{2} (x) - 7 sin (x) = 0

Let:

sin (x) = u

u^{2} - 7 u = 0

u^{2} - 7 u = 0 : u = 7, u = 0

u^{2} - 7 u = 0

Solve with the quadratic formula

u^{2} - 7 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 7, c = 0

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 0 = 7

(- 7)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

Apply radical rule:

assuming

a \geq 0

= 7

u_{1, 2} = \frac{- ( - 7 ) \pm 7}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 7}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 7}{2 \cdot 1}

u = \frac{- ( - 7 ) + 7}{2 \cdot 1} : 7

\frac{- ( - 7 ) + 7}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 + 7}{2 \cdot 1}

Add the numbers:

7 + 7 = 14

= \frac{14}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{14}{2}

Divide the numbers:

\frac{14}{2} = 7

= 7

u = \frac{- ( - 7 ) - 7}{2 \cdot 1} : 0

\frac{- ( - 7 ) - 7}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 - 7}{2 \cdot 1}

Subtract the numbers:

7 - 7 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 7, u = 0

Substitute back

u = sin (x)

sin (x) = 7, sin (x) = 0

sin (x) = 7, sin (x) = 0

sin (x) = 7 :

No Solution

sin (x) = 7

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

sin(x)=(48sin(69))/(47.5)

sin (x) = \frac{48 sin ( 6 9 ^{\circ} )}{47.5}

2cos^2(3x)+cos(3x)-1=0

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

sin(3x)=(-3)/4

sin (3 x) = \frac{- 3}{4}

4cos(x-45)+7=10

4 cos (x - 4 5^{\circ}) + 7 = 10

cos^2(x)=3(1-sin(x))

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

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-7sin(x)=0 ?
The general solution for sin^2(x)-7sin(x)=0 is x=2pin,x=pi+2pin