What is the general solution for 9sin^2(x)-6sin(x)+1=0 ?

The general solution for 9sin^2(x)-6sin(x)+1=0 is x=0.33983…+2pin,x=pi-0.33983…+2pin

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9sin^2(x)-6sin(x)+1=0

9 sin^{2} (x) - 6 sin (x) + 1 = 0

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn

Degrees

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

9 sin^{2} (x) - 6 sin (x) + 1 = 0

Solve by substitution

9 sin^{2} (x) - 6 sin (x) + 1 = 0

Let:

sin (x) = u

9 u^{2} - 6 u + 1 = 0

9 u^{2} - 6 u + 1 = 0 : u = \frac{1}{3}

9 u^{2} - 6 u + 1 = 0

Solve with the quadratic formula

9 u^{2} - 6 u + 1 = 0

Quadratic Equation Formula:

For

a = 9, b = - 6, c = 1

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

(- 6)^{2} - 4 \cdot 9 \cdot 1 = 0

(- 6)^{2} - 4 \cdot 9 \cdot 1

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 9 \cdot 1

Multiply the numbers:

4 \cdot 9 \cdot 1 = 36

= 6^{2} - 36

6^{2} = 36

= 36 - 36

Subtract the numbers:

36 - 36 = 0

= 0

u_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 \cdot 9}

u = \frac{- ( - 6 )}{2 \cdot 9}

\frac{- ( - 6 )}{2 \cdot 9} = \frac{1}{3}

\frac{- ( - 6 )}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{6}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{6}{18}

Cancel the common factor:

6

= \frac{1}{3}

u = \frac{1}{3}

The solution to the quadratic equation is:

u = \frac{1}{3}

Substitute back

u = sin (x)

sin (x) = \frac{1}{3}

sin (x) = \frac{1}{3}

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

sin (x) = \frac{1}{3}

Apply trig inverse properties

sin (x) = \frac{1}{3}

General solutions for

sin (x) = \frac{1}{3}

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

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

Combine all the solutions

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn

y, ln (x) = arctan (\frac{y}{x})

cos (a) + 1 = 4 cos (a) + 1

6 cos (x) + 3 sin (x) = 5

(2 sin (x) - 1) cos (x) = 0

sin (2 x) + cos (x) = 0, x \leq 2 π, 0

What is the general solution for 9sin^2(x)-6sin(x)+1=0 ?
The general solution for 9sin^2(x)-6sin(x)+1=0 is x=0.33983…+2pin,x=pi-0.33983…+2pin