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Popular >

11-12cos^2(θ)=0

Solution

11 - 12 cos^{2} (θ) = 0

Solution

θ = 0.29284 \dots + 2 πn, θ = 2 π - 0.29284 \dots + 2 πn, θ = 2.84874 \dots + 2 πn, θ = - 2.84874 \dots + 2 πn

Degrees

θ = 16.77865 \dots^{\circ} + 36 0^{\circ} n, θ = 343.22134 \dots^{\circ} + 36 0^{\circ} n, θ = 163.22134 \dots^{\circ} + 36 0^{\circ} n, θ = - 163.22134 \dots^{\circ} + 36 0^{\circ} n

Solution steps

11 - 12 cos^{2} (θ) = 0

Solve by substitution

11 - 12 cos^{2} (θ) = 0

Let:

cos (θ) = u

11 - 12 u^{2} = 0

11 - 12 u^{2} = 0 : u = \frac{33}{6}, u = - \frac{33}{6}

11 - 12 u^{2} = 0

Move

11

to the right side

11 - 12 u^{2} = 0

Subtract

11

from both sides

11 - 12 u^{2} - 11 = 0 - 11

Simplify

- 12 u^{2} = - 11

- 12 u^{2} = - 11

Divide both sides by

- 12

- 12 u^{2} = - 11

Divide both sides by

- 12

\frac{- 12 u ^{2}}{- 12} = \frac{- 11}{- 12}

Simplify

u^{2} = \frac{11}{12}

u^{2} = \frac{11}{12}

For

x^{2} = f (a)

the solutions are

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

u = \frac{11}{12}, u = - \frac{11}{12}

\frac{11}{12} = \frac{33}{6}

\frac{11}{12}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{11}{12}

12 = 23

12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= \frac{11}{2 3}

Rationalize

\frac{11}{2 3} : \frac{33}{6}

\frac{11}{2 3}

Multiply by the conjugate

\frac{3}{3}

= \frac{11 3}{2 3 3}

113 = 33

113

Apply radical rule:

a b = a \cdot b

113 = 11 \cdot 3

= 11 \cdot 3

Multiply the numbers:

11 \cdot 3 = 33

= 33

233 = 6

233

Apply radical rule:

a a = a

33 = 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{33}{6}

= \frac{33}{6}

- \frac{11}{12} = - \frac{33}{6}

- \frac{11}{12}

Simplify

\frac{11}{12} : \frac{11}{2 3}

\frac{11}{12}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{11}{12}

12 = 23

12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= \frac{11}{2 3}

= - \frac{11}{2 3}

Rationalize

- \frac{11}{2 3} : - \frac{33}{6}

- \frac{11}{2 3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{11 3}{2 3 3}

113 = 33

113

Apply radical rule:

a b = a \cdot b

113 = 11 \cdot 3

= 11 \cdot 3

Multiply the numbers:

11 \cdot 3 = 33

= 33

233 = 6

233

Apply radical rule:

a a = a

33 = 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= - \frac{33}{6}

= - \frac{33}{6}

u = \frac{33}{6}, u = - \frac{33}{6}

Substitute back

u = cos (θ)

cos (θ) = \frac{33}{6}, cos (θ) = - \frac{33}{6}

cos (θ) = \frac{33}{6}, cos (θ) = - \frac{33}{6}

cos (θ) = \frac{33}{6} : θ = arccos (\frac{33}{6}) + 2 πn, θ = 2 π - arccos (\frac{33}{6}) + 2 πn

cos (θ) = \frac{33}{6}

Apply trig inverse properties

cos (θ) = \frac{33}{6}

General solutions for

cos (θ) = \frac{33}{6}

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

θ = arccos (\frac{33}{6}) + 2 πn, θ = 2 π - arccos (\frac{33}{6}) + 2 πn

θ = arccos (\frac{33}{6}) + 2 πn, θ = 2 π - arccos (\frac{33}{6}) + 2 πn

cos (θ) = - \frac{33}{6} : θ = arccos (- \frac{33}{6}) + 2 πn, θ = - arccos (- \frac{33}{6}) + 2 πn

cos (θ) = - \frac{33}{6}

Apply trig inverse properties

cos (θ) = - \frac{33}{6}

General solutions for

cos (θ) = - \frac{33}{6}

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

θ = arccos (- \frac{33}{6}) + 2 πn, θ = - arccos (- \frac{33}{6}) + 2 πn

θ = arccos (- \frac{33}{6}) + 2 πn, θ = - arccos (- \frac{33}{6}) + 2 πn

Combine all the solutions

θ = arccos (\frac{33}{6}) + 2 πn, θ = 2 π - arccos (\frac{33}{6}) + 2 πn, θ = arccos (- \frac{33}{6}) + 2 πn, θ = - arccos (- \frac{33}{6}) + 2 πn

Show solutions in decimal form

θ = 0.29284 \dots + 2 πn, θ = 2 π - 0.29284 \dots + 2 πn, θ = 2.84874 \dots + 2 πn, θ = - 2.84874 \dots + 2 πn

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