English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

81(cos^2(x))=6

Solution

81 (cos^{2} (x)) = 6

Solution

x = 1.29515 \dots + 2 πn, x = 2 π - 1.29515 \dots + 2 πn, x = 1.84643 \dots + 2 πn, x = - 1.84643 \dots + 2 πn

Degrees

x = 74.20683 \dots^{\circ} + 36 0^{\circ} n, x = 285.79316 \dots^{\circ} + 36 0^{\circ} n, x = 105.79316 \dots^{\circ} + 36 0^{\circ} n, x = - 105.79316 \dots^{\circ} + 36 0^{\circ} n

Solution steps

81 (cos^{2} (x)) = 6

Solve by substitution

81 cos^{2} (x) = 6

Let:

cos (x) = u

81 u^{2} = 6

81 u^{2} = 6 : u = \frac{6}{9}, u = - \frac{6}{9}

81 u^{2} = 6

Divide both sides by

81

81 u^{2} = 6

Divide both sides by

81

\frac{81 u ^{2}}{81} = \frac{6}{81}

Simplify

u^{2} = \frac{2}{27}

u^{2} = \frac{2}{27}

For

x^{2} = f (a)

the solutions are

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

u = \frac{2}{27}, u = - \frac{2}{27}

\frac{2}{27} = \frac{6}{9}

\frac{2}{27}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2}{27}

27 = 33

27

Prime factorization of

27 : 3^{3}

27

27

divides by

327 = 9 \cdot 3

= 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3 \cdot 3

3

is a prime number, therefore no further factorization is possible

= 3 \cdot 3 \cdot 3

= 3^{3}

= 3^{3}

Apply exponent rule:

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

= 3^{2} \cdot 3

Apply radical rule:

= 3 3^{2}

Apply radical rule:

3^{2} = 3

= 33

= \frac{2}{3 3}

Rationalize

\frac{2}{3 3} : \frac{6}{9}

\frac{2}{3 3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3 3}

23 = 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

333 = 9

333

Apply exponent rule:

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

333 = 3 \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} = 3^{1 + \frac{1}{2} + \frac{1}{2}}

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

Add similar elements:

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

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

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

2 \cdot \frac{1}{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3^{2}

3^{2} = 9

= 9

= \frac{6}{9}

= \frac{6}{9}

- \frac{2}{27} = - \frac{6}{9}

- \frac{2}{27}

Simplify

\frac{2}{27} : \frac{2}{3 3}

\frac{2}{27}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2}{27}

27 = 33

27

Prime factorization of

27 : 3^{3}

27

27

divides by

327 = 9 \cdot 3

= 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3 \cdot 3

3

is a prime number, therefore no further factorization is possible

= 3 \cdot 3 \cdot 3

= 3^{3}

= 3^{3}

Apply exponent rule:

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

= 3^{2} \cdot 3

Apply radical rule:

= 3 3^{2}

Apply radical rule:

3^{2} = 3

= 33

= \frac{2}{3 3}

= - \frac{2}{3 3}

Rationalize

- \frac{2}{3 3} : - \frac{6}{9}

- \frac{2}{3 3}

Multiply by the conjugate

\frac{3}{3}

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

23 = 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

333 = 9

333

Apply exponent rule:

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

333 = 3 \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} = 3^{1 + \frac{1}{2} + \frac{1}{2}}

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

Add similar elements:

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

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

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

2 \cdot \frac{1}{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3^{2}

3^{2} = 9

= 9

= - \frac{6}{9}

= - \frac{6}{9}

u = \frac{6}{9}, u = - \frac{6}{9}

Substitute back

u = cos (x)

cos (x) = \frac{6}{9}, cos (x) = - \frac{6}{9}

cos (x) = \frac{6}{9}, cos (x) = - \frac{6}{9}

cos (x) = \frac{6}{9} : x = arccos (\frac{6}{9}) + 2 πn, x = 2 π - arccos (\frac{6}{9}) + 2 πn

cos (x) = \frac{6}{9}

Apply trig inverse properties

cos (x) = \frac{6}{9}

General solutions for

cos (x) = \frac{6}{9}

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

x = arccos (\frac{6}{9}) + 2 πn, x = 2 π - arccos (\frac{6}{9}) + 2 πn

x = arccos (\frac{6}{9}) + 2 πn, x = 2 π - arccos (\frac{6}{9}) + 2 πn

cos (x) = - \frac{6}{9} : x = arccos (- \frac{6}{9}) + 2 πn, x = - arccos (- \frac{6}{9}) + 2 πn

cos (x) = - \frac{6}{9}

Apply trig inverse properties

cos (x) = - \frac{6}{9}

General solutions for

cos (x) = - \frac{6}{9}

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

x = arccos (- \frac{6}{9}) + 2 πn, x = - arccos (- \frac{6}{9}) + 2 πn

x = arccos (- \frac{6}{9}) + 2 πn, x = - arccos (- \frac{6}{9}) + 2 πn

Combine all the solutions

x = arccos (\frac{6}{9}) + 2 πn, x = 2 π - arccos (\frac{6}{9}) + 2 πn, x = arccos (- \frac{6}{9}) + 2 πn, x = - arccos (- \frac{6}{9}) + 2 πn

Show solutions in decimal form

x = 1.29515 \dots + 2 πn, x = 2 π - 1.29515 \dots + 2 πn, x = 1.84643 \dots + 2 πn, x = - 1.84643 \dots + 2 πn

Popular Examples

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

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

6cos^2(x)+sin(x)-5=0

6 cos^{2} (x) + sin (x) - 5 = 0

tan(θ)= 23/72

tan (θ) = \frac{23}{72}

sin(θ-pi/2)=cos(θ)

sin (θ - \frac{π}{2}) = cos (θ)

sec^2(x)-4/3 =0

sec^{2} (x) - \frac{4}{3} = 0