What is the general solution for 64cos^2(θ)=49,(0<= ,θ<= 360) ?

Q: What is the general solution for 64cos^2(θ)=49,(0<= ,θ<= 360) ?

The general solution for 64cos^2(θ)=49,(0<= ,θ<= 360) is No Solution for θ\in\mathbb{R}

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64cos^2(θ)=49,(0<= ,θ<= 360)

Solution

64 cos^{2} (θ) = 49, (0^{\circ} \leq, θ \leq 36 0^{\circ})

Solution

No Solution for θ \in R

Solution steps

64 cos^{2} (θ) = 49, (0^{\circ} \leq, θ \leq 36 0^{\circ})

Solve by substitution

64 cos^{2} (θ) = 49

Let:

cos (θ) = u

64 u^{2} = 49

64 u^{2} = 49 : u = \frac{7}{8}, u = - \frac{7}{8}

64 u^{2} = 49

Divide both sides by

64

64 u^{2} = 49

Divide both sides by

64

\frac{64 u ^{2}}{64} = \frac{49}{64}

Simplify

u^{2} = \frac{49}{64}

u^{2} = \frac{49}{64}

For

x^{2} = f (a)

the solutions are

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

u = \frac{49}{64}, u = - \frac{49}{64}

\frac{49}{64} = \frac{7}{8}

\frac{49}{64}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{49}{64}

64 = 8

64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

= \frac{49}{8}

49 = 7

49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

= \frac{7}{8}

- \frac{49}{64} = - \frac{7}{8}

- \frac{49}{64}

Simplify

\frac{49}{64} : \frac{7}{8}

\frac{49}{64}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{49}{64}

64 = 8

64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

= \frac{49}{8}

49 = 7

49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

= \frac{7}{8}

= - \frac{7}{8}

u = \frac{7}{8}, u = - \frac{7}{8}

Substitute back

u = cos (θ)

cos (θ) = \frac{7}{8}, cos (θ) = - \frac{7}{8}

cos (θ) = \frac{7}{8}, cos (θ) = - \frac{7}{8}

cos (θ) = \frac{7}{8}, (0 \leq, θ \leq 36 0^{\circ}) :

No Solution

cos (θ) = \frac{7}{8}, (0 \leq, θ \leq 36 0^{\circ})

Apply trig inverse properties

cos (θ) = \frac{7}{8}

General solutions for

cos (θ) = \frac{7}{8}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{7}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{7}{8}) + 36 0^{\circ} n

θ = arccos (\frac{7}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{7}{8}) + 36 0^{\circ} n

Solutions for the range

(0 \leq, θ \leq 36 0^{\circ})

No Solution

cos (θ) = - \frac{7}{8}, (0 \leq, θ \leq 36 0^{\circ}) :

No Solution

cos (θ) = - \frac{7}{8}, (0 \leq, θ \leq 36 0^{\circ})

Apply trig inverse properties

cos (θ) = - \frac{7}{8}

General solutions for

cos (θ) = - \frac{7}{8}

cos (x) = - a \Rightarrow x = arccos (- a) + 36 0^{\circ} n, x = - arccos (- a) + 36 0^{\circ} n

θ = arccos (- \frac{7}{8}) + 36 0^{\circ} n, θ = - arccos (- \frac{7}{8}) + 36 0^{\circ} n

θ = arccos (- \frac{7}{8}) + 36 0^{\circ} n, θ = - arccos (- \frac{7}{8}) + 36 0^{\circ} n

Solutions for the range

(0 \leq, θ \leq 36 0^{\circ})

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 64cos^2(θ)=49,(0<= ,θ<= 360) ?
The general solution for 64cos^2(θ)=49,(0<= ,θ<= 360) is No Solution for θ\in\mathbb{R}