What is the general solution for 1/(cos^2(x))+(1/(2cos(x)))=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/(cos^2(x))+(1/(2cos(x)))=0

Solution

\frac{1}{cos ^{2} ( x )} + (\frac{1}{2 cos ( x )}) = 0

Solution

No Solution for x \in R

Solution steps

\frac{1}{cos ^{2} ( x )} + (\frac{1}{2 cos ( x )}) = 0

Solve by substitution

\frac{1}{cos ^{2} ( x )} + \frac{1}{2 cos ( x )} = 0

Let:

cos (x) = u

\frac{1}{u ^{2}} + \frac{1}{2 u} = 0

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

\frac{1}{u ^{2}} + \frac{1}{2 u} = 0

Multiply by LCM

\frac{1}{u ^{2}} + \frac{1}{2 u} = 0

Find Least Common Multiplier of

u^{2}, 2 u : 2 u^{2}

u^{2}, 2 u

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

u^{2}

2 u

= 2 u^{2}

Multiply by LCM=

2 u^{2}

\frac{1}{u ^{2}} \cdot 2 u^{2} + \frac{1}{2 u} \cdot 2 u^{2} = 0 \cdot 2 u^{2}

Simplify

\frac{1}{u ^{2}} \cdot 2 u^{2} + \frac{1}{2 u} \cdot 2 u^{2} = 0 \cdot 2 u^{2}

Simplify

\frac{1}{u ^{2}} \cdot 2 u^{2} : 2

\frac{1}{u ^{2}} \cdot 2 u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 u ^{2}}{u ^{2}}

Cancel the common factor:

u^{2}

= 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

Simplify

\frac{1}{2 u} \cdot 2 u^{2} : u

\frac{1}{2 u} \cdot 2 u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 u ^{2}}{2 u}

Cancel the common factor:

2

= \frac{1 \cdot u ^{2}}{u}

Multiply:

1 \cdot u^{2} = u^{2}

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

Cancel the common factor:

u

= u

Simplify

0 \cdot 2 u^{2} : 0

0 \cdot 2 u^{2}

Apply rule

0 \cdot a = 0

= 0

2 + u = 0

2 + u = 0

2 + u = 0

Move

2

to the right side

2 + u = 0

Subtract

2

from both sides

2 + u - 2 = 0 - 2

Simplify

u = - 2

u = - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{1}{u ^{2}} + \frac{1}{2 u}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Solve

2 u = 0 : u = 0

2 u = 0

Divide both sides by

2

2 u = 0

Divide both sides by

2

\frac{2 u}{2} = \frac{0}{2}

Simplify

u = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 2

Substitute back

u = cos (x)

cos (x) = - 2

cos (x) = - 2

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

sqrt(3)sin(x)+cos(x)=sqrt(2)

3 sin (x) + cos (x) = 2

cos(α)=sqrt((1+1/5)/2)

cos (α) = \frac{1 + \frac{1}{5}}{2}

(tan(x))(sin^2(x)-a)(sec(x)-a)=0,0<a<1

(tan (x)) (sin^{2} (x) - a) (sec (x) - a) = 0, 0 < a < 1

sin(pi/2 (x+1))= 2/5

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

sin(*+pi/2)=cos(x)

sin (\cdot + \frac{π}{2}) = cos (x)

Frequently Asked Questions (FAQ)

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