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Popular >

cos^2(β)= 16/25 ,0<= β<= 2pi

Solution

cos^{2} (β) = \frac{16}{25}, 0 \leq β \leq 2 π

Solution

β = 0.64350 \dots, β = 2 π - 0.64350 \dots, β = 2.49809 \dots, β = - 2.49809 \dots + 2 π

+1

Degrees

β = 36.86989 \dots^{\circ}, β = 323.13010 \dots^{\circ}, β = 143.13010 \dots^{\circ}, β = 216.86989 \dots^{\circ}

Solution steps

cos^{2} (β) = \frac{16}{25}, 0 \leq β \leq 2 π

Solve by substitution

cos^{2} (β) = \frac{16}{25}

Let:

cos (β) = u

u^{2} = \frac{16}{25}

u^{2} = \frac{16}{25} : u = \frac{4}{5}, u = - \frac{4}{5}

u^{2} = \frac{16}{25}

For

x^{2} = f (a)

the solutions are

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

u = \frac{16}{25}, u = - \frac{16}{25}

\frac{16}{25} = \frac{4}{5}

\frac{16}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{16}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{16}{5}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{4}{5}

- \frac{16}{25} = - \frac{4}{5}

- \frac{16}{25}

Simplify

\frac{16}{25} : \frac{4}{5}

\frac{16}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{16}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{16}{5}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{4}{5}

= - \frac{4}{5}

u = \frac{4}{5}, u = - \frac{4}{5}

Substitute back

u = cos (β)

cos (β) = \frac{4}{5}, cos (β) = - \frac{4}{5}

cos (β) = \frac{4}{5}, cos (β) = - \frac{4}{5}

cos (β) = \frac{4}{5}, 0 \leq β \leq 2 π : β = arccos (\frac{4}{5}), β = 2 π - arccos (\frac{4}{5})

cos (β) = \frac{4}{5}, 0 \leq β \leq 2 π

Apply trig inverse properties

cos (β) = \frac{4}{5}

General solutions for

cos (β) = \frac{4}{5}

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

β = arccos (\frac{4}{5}) + 2 πn, β = 2 π - arccos (\frac{4}{5}) + 2 πn

β = arccos (\frac{4}{5}) + 2 πn, β = 2 π - arccos (\frac{4}{5}) + 2 πn

Solutions for the range

0 \leq β \leq 2 π

β = arccos (\frac{4}{5}), β = 2 π - arccos (\frac{4}{5})

cos (β) = - \frac{4}{5}, 0 \leq β \leq 2 π : β = arccos (- \frac{4}{5}), β = - arccos (- \frac{4}{5}) + 2 π

cos (β) = - \frac{4}{5}, 0 \leq β \leq 2 π

Apply trig inverse properties

cos (β) = - \frac{4}{5}

General solutions for

cos (β) = - \frac{4}{5}

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

β = arccos (- \frac{4}{5}) + 2 πn, β = - arccos (- \frac{4}{5}) + 2 πn

β = arccos (- \frac{4}{5}) + 2 πn, β = - arccos (- \frac{4}{5}) + 2 πn

Solutions for the range

0 \leq β \leq 2 π

β = arccos (- \frac{4}{5}), β = - arccos (- \frac{4}{5}) + 2 π

Combine all the solutions

β = arccos (\frac{4}{5}), β = 2 π - arccos (\frac{4}{5}), β = arccos (- \frac{4}{5}), β = - arccos (- \frac{4}{5}) + 2 π

Show solutions in decimal form

β = 0.64350 \dots, β = 2 π - 0.64350 \dots, β = 2.49809 \dots, β = - 2.49809 \dots + 2 π

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