What is the general solution for (sin^2(x))/(1-cos(x))=1.23 ?

The general solution for (sin^2(x))/(1-cos(x))=1.23 is x=1.33871…+2pin,x=2pi-1.33871…+2pin

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(sin^2(x))/(1-cos(x))=1.23

Solution

\frac{sin ^{2} ( x )}{1 - cos ( x )} = 1.23

Solution

x = 1.33871 \dots + 2 πn, x = 2 π - 1.33871 \dots + 2 πn

Degrees

x = 76.70292 \dots^{\circ} + 36 0^{\circ} n, x = 283.29707 \dots^{\circ} + 36 0^{\circ} n

Solution steps

\frac{sin ^{2} ( x )}{1 - cos ( x )} = 1.23

Subtract

1.23

from both sides

\frac{sin ^{2} ( x )}{1 - cos ( x )} - 1.23 = 0

Rewrite using trig identities

- 1.23 + \frac{sin ^{2} ( x )}{1 - cos ( x )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 1.23 + \frac{1 - cos ^{2} ( x )}{1 - cos ( x )}

Simplify

- 1.23 + \frac{1 - cos ^{2} ( x )}{1 - cos ( x )} : cos (x) - 0.23

- 1.23 + \frac{1 - cos ^{2} ( x )}{1 - cos ( x )}

Cancel

\frac{1 - cos ^{2} ( x )}{1 - cos ( x )} : cos (x) + 1

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

Factor

1 - cos^{2} (x) : - (cos (x) + 1) (cos (x) - 1)

1 - cos^{2} (x)

Factor out common term

- 1

= - (cos^{2} (x) - 1)

Factor

cos^{2} (x) - 1 : (cos (x) + 1) (cos (x) - 1)

cos^{2} (x) - 1

Rewrite

1

1^{2}

= cos^{2} (x) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

= - \frac{( cos ( x ) + 1 ) ( cos ( x ) - 1 )}{1 - cos ( x )}

Cancel

- \frac{( cos ( x ) + 1 ) ( cos ( x ) - 1 )}{1 - cos ( x )} : cos (x) + 1

- \frac{( cos ( x ) + 1 ) ( cos ( x ) - 1 )}{1 - cos ( x )}

- cos (x) + 1 = - (cos (x) - 1)

= - \frac{( cos ( x ) + 1 ) ( cos ( x ) - 1 )}{- ( cos ( x ) - 1 )}

Refine

= \frac{( cos ( x ) + 1 ) ( cos ( x ) - 1 )}{cos ( x ) - 1}

Cancel the common factor:

cos (x) - 1

= cos (x) + 1

= cos (x) + 1

= - 1.23 + cos (x) + 1

Group like terms

= cos (x) - 1.23 + 1

Add/Subtract the numbers:

- 1.23 + 1 = - 0.23

= cos (x) - 0.23

= cos (x) - 0.23

- 0.23 + cos (x) = 0

Move

0.23

to the right side

- 0.23 + cos (x) = 0

Add

0.23

to both sides

- 0.23 + cos (x) + 0.23 = 0 + 0.23

Simplify

cos (x) = 0.23

cos (x) = 0.23

Apply trig inverse properties

cos (x) = 0.23

General solutions for

cos (x) = 0.23

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

x = arccos (0.23) + 2 πn, x = 2 π - arccos (0.23) + 2 πn

x = arccos (0.23) + 2 πn, x = 2 π - arccos (0.23) + 2 πn

Show solutions in decimal form

x = 1.33871 \dots + 2 πn, x = 2 π - 1.33871 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (sin^2(x))/(1-cos(x))=1.23 ?
The general solution for (sin^2(x))/(1-cos(x))=1.23 is x=1.33871…+2pin,x=2pi-1.33871…+2pin