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Popular >

9sin^2(x)-1=0

Solution

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

Solution

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

+1

Degrees

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n, x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

9 u^{2} - 1 = 0

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

9 u^{2} - 1 = 0

Move

1

to the right side

9 u^{2} - 1 = 0

Add

1

to both sides

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

Simplify

9 u^{2} = 1

9 u^{2} = 1

Divide both sides by

9

9 u^{2} = 1

Divide both sides by

9

\frac{9 u ^{2}}{9} = \frac{1}{9}

Simplify

u^{2} = \frac{1}{9}

u^{2} = \frac{1}{9}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{9}, u = - \frac{1}{9}

\frac{1}{9} = \frac{1}{3}

\frac{1}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{1}{3}

Apply rule

1 = 1

= \frac{1}{3}

- \frac{1}{9} = - \frac{1}{3}

- \frac{1}{9}

Simplify

\frac{1}{9} : \frac{1}{3}

\frac{1}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{1}{3}

= - \frac{1}{3}

Apply rule

1 = 1

= - \frac{1}{3}

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = sin (x)

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

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

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

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