What is the general solution for solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 ?

The general solution for solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 is No Solution for x\in\mathbb{R}

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Popular >

solvefor x,cos^2(x)z^2-2cos^2(x)z+1=0

Solution

solve for

x, cos^{2} (x) \cdot z^{2} - 2 cos^{2} (x) \cdot z + 1 = 0

Solution

No Solution for x \in R

Solution steps

cos^{2} (x) z^{2} - 2 cos^{2} (x) z + 1 = 0

Solve by substitution

cos^{2} (x) z^{2} - 2 cos^{2} (x) z + 1 = 0

Let:

cos (x) = u

u^{2} z^{2} - 2 u^{2} z + 1 = 0

u^{2} z^{2} - 2 u^{2} z + 1 = 0 : u = i \frac{1}{z ^{2} - 2 z}, u = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

u^{2} z^{2} - 2 u^{2} z + 1 = 0

Move

1

to the right side

u^{2} z^{2} - 2 u^{2} z + 1 = 0

Subtract

1

from both sides

u^{2} z^{2} - 2 u^{2} z + 1 - 1 = 0 - 1

Simplify

u^{2} z^{2} - 2 u^{2} z = - 1

u^{2} z^{2} - 2 u^{2} z = - 1

Factor

u^{2} z^{2} - 2 u^{2} z : z u^{2} (z - 2)

u^{2} z^{2} - 2 u^{2} z

Apply exponent rule:

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

z^{2} = zz

= u^{2} zz - 2 u^{2} z

Factor out common term

u^{2} z

= u^{2} z (z - 2)

z u^{2} (z - 2) = - 1

Divide both sides by

z (z - 2); z \neq = 0, z \neq = 2

z u^{2} (z - 2) = - 1

Divide both sides by

z (z - 2); z \neq = 0, z \neq = 2

\frac{z u ^{2} ( z - 2 )}{z ( z - 2 )} = \frac{- 1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

Simplify

u^{2} = - \frac{1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

u^{2} = - \frac{1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

For

x^{2} = f (a)

the solutions are

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

u = - \frac{1}{z ( z - 2 )}, u = - - \frac{1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

Simplify

- \frac{1}{z ( z - 2 )} : i \frac{1}{z ^{2} - 2 z}

- \frac{1}{z ( z - 2 )}

Apply radical rule:

- a = - 1 a,

assuming

a \geq 0

- \frac{1}{z ( z - 2 )} = - 1 \frac{1}{z ( z - 2 )}

= - 1 \frac{1}{z ( z - 2 )}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{z ( z - 2 )}

Rewrite

i \frac{1}{z ( z - 2 )}

in standard complex form:

\frac{1}{z ^{2} - 2 z} i

i \frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

Expand

z (z - 2) : z^{2} - 2 z

z (z - 2)

Apply the distributive law:

a (b - c) = ab - a c

a = z, b = z, c = 2

= zz - z \cdot 2

= zz - 2 z

zz = z^{2}

zz

Apply exponent rule:

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

zz = z^{1 + 1}

= z^{1 + 1}

Add the numbers:

1 + 1 = 2

= z^{2}

= z^{2} - 2 z

= \frac{1}{z ^{2} - 2 z}

= \frac{1}{z ^{2} - 2 z}

= i \frac{1}{z ^{2} - 2 z}

= \frac{1}{z ^{2} - 2 z} i

Simplify

- - \frac{1}{z ( z - 2 )} : - i \frac{1}{z ^{2} - 2 z}

- - \frac{1}{z ( z - 2 )}

Simplify

- \frac{1}{z ( z - 2 )} : i \frac{1}{z ( z - 2 )}

- \frac{1}{z ( z - 2 )}

Apply radical rule:

- a = - 1 a,

assuming

a \geq 0

- \frac{1}{z ( z - 2 )} = - 1 \frac{1}{z ( z - 2 )}

= - 1 \frac{1}{z ( z - 2 )}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{z ( z - 2 )}

= - i \frac{1}{z ( z - 2 )}

Rewrite

- i \frac{1}{z ( z - 2 )}

in standard complex form:

- \frac{1}{z ^{2} - 2 z} i

- i \frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

Expand

z (z - 2) : z^{2} - 2 z

z (z - 2)

Apply the distributive law:

a (b - c) = ab - a c

a = z, b = z, c = 2

= zz - z \cdot 2

= zz - 2 z

zz = z^{2}

zz

Apply exponent rule:

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

zz = z^{1 + 1}

= z^{1 + 1}

Add the numbers:

1 + 1 = 2

= z^{2}

= z^{2} - 2 z

= \frac{1}{z ^{2} - 2 z}

= \frac{1}{z ^{2} - 2 z}

= - i \frac{1}{z ^{2} - 2 z}

= - \frac{1}{z ^{2} - 2 z} i

u = i \frac{1}{z ^{2} - 2 z}, u = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

Substitute back

u = cos (x)

cos (x) = i \frac{1}{z ^{2} - 2 z}, cos (x) = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

cos (x) = i \frac{1}{z ^{2} - 2 z}, cos (x) = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

cos (x) = i \frac{1}{z ^{2} - 2 z} :

No Solution

cos (x) = i \frac{1}{z ^{2} - 2 z}

No Solution

cos (x) = - i \frac{1}{z ^{2} - 2 z} :

No Solution

cos (x) = - i \frac{1}{z ^{2} - 2 z}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 ?
The general solution for solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 is No Solution for x\in\mathbb{R}