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Popular >

1/9+cos^2(x)=1

Solution

\frac{1}{9} + cos^{2} (x) = 1

Solution

x = 0.33983 \dots + 2 πn, x = 2 π - 0.33983 \dots + 2 πn, x = 2.80175 \dots + 2 πn, x = - 2.80175 \dots + 2 πn

Degrees

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 340.52877 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n, x = - 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1}{9} + cos^{2} (x) = 1

Solve by substitution

\frac{1}{9} + cos^{2} (x) = 1

Let:

cos (x) = u

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

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

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

Move

\frac{1}{9}

to the right side

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

Subtract

\frac{1}{9}

from both sides

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

Simplify

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

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

Simplify

1 - \frac{1}{9} : \frac{8}{9}

1 - \frac{1}{9}

Convert element to fraction:

1 = \frac{1 \cdot 9}{9}

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 9 - 1 = 8

1 \cdot 9 - 1

Multiply the numbers:

1 \cdot 9 = 9

= 9 - 1

Subtract the numbers:

9 - 1 = 8

= 8

= \frac{8}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{8}{9}, u = - \frac{8}{9}

\frac{8}{9} = \frac{2 2}{3}

\frac{8}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{8}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{8}{3}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= \frac{2 2}{3}

- \frac{8}{9} = - \frac{2 2}{3}

- \frac{8}{9}

Simplify

\frac{8}{9} : \frac{2 2}{3}

\frac{8}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{8}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{8}{3}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= \frac{2 2}{3}

= - \frac{2 2}{3}

u = \frac{2 2}{3}, u = - \frac{2 2}{3}

Substitute back

u = cos (x)

cos (x) = \frac{2 2}{3}, cos (x) = - \frac{2 2}{3}

cos (x) = \frac{2 2}{3}, cos (x) = - \frac{2 2}{3}

cos (x) = \frac{2 2}{3} : x = arccos (\frac{2 2}{3}) + 2 πn, x = 2 π - arccos (\frac{2 2}{3}) + 2 πn

cos (x) = \frac{2 2}{3}

Apply trig inverse properties

cos (x) = \frac{2 2}{3}

General solutions for

cos (x) = \frac{2 2}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

cos (x) = - \frac{2 2}{3} : x = arccos (- \frac{2 2}{3}) + 2 πn, x = - arccos (- \frac{2 2}{3}) + 2 πn

cos (x) = - \frac{2 2}{3}

Apply trig inverse properties

cos (x) = - \frac{2 2}{3}

General solutions for

cos (x) = - \frac{2 2}{3}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{2 2}{3}) + 2 πn, x = - arccos (- \frac{2 2}{3}) + 2 πn

x = arccos (- \frac{2 2}{3}) + 2 πn, x = - arccos (- \frac{2 2}{3}) + 2 πn

Combine all the solutions

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

Show solutions in decimal form

x = 0.33983 \dots + 2 πn, x = 2 π - 0.33983 \dots + 2 πn, x = 2.80175 \dots + 2 πn, x = - 2.80175 \dots + 2 πn

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