What is the general solution for (a^{0.2})/((cos^2(x))-cos^2(x)-1)=0 ?

The general solution for (a^{0.2})/((cos^2(x))-cos^2(x)-1)=0 is No Solution for x\in\mathbb{R}

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(a^{0.2})/((cos^2(x))-cos^2(x)-1)=0

\frac{a ^{0.2}}{( cos ^{2} ( x ) ) - cos ^{2} ( x ) - 1} = 0

No Solution for x \in R

Solution steps

\frac{a ^{0.2}}{( cos ^{2} ( x ) ) - cos ^{2} ( x ) - 1} = 0

Solve by substitution

\frac{a ^{0.2}}{cos ^{2} ( x ) - cos ^{2} ( x ) - 1} = 0

Let:

cos (x) = u

\frac{a ^{0.2}}{u ^{2} - u ^{2} - 1} = 0

\frac{a ^{0.2}}{u ^{2} - u ^{2} - 1} = 0 :

True for all

u; - a^{0.2} = 0

\frac{a ^{0.2}}{u ^{2} - u ^{2} - 1} = 0

Simplify

\frac{a ^{0.2}}{u ^{2} - u ^{2} - 1} : - a^{0.2}

\frac{a ^{0.2}}{u ^{2} - u ^{2} - 1}

Add similar elements:

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

= \frac{a ^{0.2}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{a ^{0.2}}{1}

Apply rule

\frac{a}{1} = a

= - a^{0.2}

- a^{0.2} = 0

Both sides are equal

True for all u; - a^{0.2} = 0

Substitute back

u = cos (x)

True for all cos (x); - a^{0.2} = 0

True for all cos (x); - a^{0.2} = 0

cos (x) =

True for all

u : x = arccos (True for all u) + 2 πn, x = - arccos (True for all u) + 2 πn

cos (x) = True for all u

Apply trig inverse properties

cos (x) = True for all u

General solutions for

cos (x) =

True for all

u

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

x = arccos (True for all u) + 2 πn, x = - arccos (True for all u) + 2 πn

x = arccos (True for all u) + 2 πn, x = - arccos (True for all u) + 2 πn

Combine all the solutions

x = arccos (True for all u) + 2 πn, x = - arccos (True for all u) + 2 πn

Since the equation is undefined for:

arccos (True for all u) + 2 πn, - arccos (True for all u) + 2 πn

No Solution for x \in R

What is the general solution for (a^{0.2})/((cos^2(x))-cos^2(x)-1)=0 ?
The general solution for (a^{0.2})/((cos^2(x))-cos^2(x)-1)=0 is No Solution for x\in\mathbb{R}