What is the general solution for solvefor t,x=acos^2(t)+bsin^2(t) ?

The general solution for solvefor t,x=acos^2(t)+bsin^2(t) is

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solvefor t,x=acos^2(t)+bsin^2(t)

Solution

solve for

t, x = a cos^{2} (t) + b sin^{2} (t)

Solution

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

Solution steps

x = a cos^{2} (t) + b sin^{2} (t)

Switch sides

a cos^{2} (t) + b sin^{2} (t) = x

Subtract

x

from both sides

a cos^{2} (t) + b sin^{2} (t) - x = 0

Rewrite using trig identities

- x + cos^{2} (t) a + sin^{2} (t) b

Use the Pythagorean identity:

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

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

= - x + (1 - sin^{2} (t)) a + sin^{2} (t) b

- x + (1 - sin^{2} (t)) a + sin^{2} (t) b = 0

Solve by substitution

- x + (1 - sin^{2} (t)) a + sin^{2} (t) b = 0

Let:

sin (t) = u

- x + (1 - u^{2}) a + u^{2} b = 0

- x + (1 - u^{2}) a + u^{2} b = 0 : u = \frac{x - a}{- a + b}, u = - \frac{x - a}{- a + b}; - a + b \neq = 0

- x + (1 - u^{2}) a + u^{2} b = 0

Move

x

to the right side

- x + (1 - u^{2}) a + u^{2} b = 0

Add

x

to both sides

- x + (1 - u^{2}) a + u^{2} b + x = 0 + x

Simplify

(1 - u^{2}) a + u^{2} b = x

(1 - u^{2}) a + u^{2} b = x

Expand

(1 - u^{2}) a + u^{2} b : a - a u^{2} + b u^{2}

(1 - u^{2}) a + u^{2} b

= a (1 - u^{2}) + b u^{2}

Expand

a (1 - u^{2}) : a - a u^{2}

a (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = a, b = 1, c = u^{2}

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

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

Multiply:

1 \cdot a = a

= a - a u^{2}

= a - a u^{2} + u^{2} b

= a - a u^{2} + b u^{2}

a - a u^{2} + b u^{2} = x

Move

a

to the right side

a - a u^{2} + b u^{2} = x

Subtract

a

from both sides

a - a u^{2} + b u^{2} - a = x - a

Simplify

- a u^{2} + b u^{2} = x - a

- a u^{2} + b u^{2} = x - a

Factor

- a u^{2} + b u^{2} : u^{2} (- a + b)

- a u^{2} + b u^{2}

Factor out common term

u^{2}

= u^{2} (- a + b)

u^{2} (- a + b) = x - a

Divide both sides by

- a + b; - a + b \neq = 0

u^{2} (- a + b) = x - a

Divide both sides by

- a + b; - a + b \neq = 0

\frac{u ^{2} ( - a + b )}{- a + b} = \frac{x}{- a + b} - \frac{a}{- a + b}; - a + b \neq = 0

Simplify

\frac{u ^{2} ( - a + b )}{- a + b} = \frac{x}{- a + b} - \frac{a}{- a + b}

Simplify

\frac{u ^{2} ( - a + b )}{- a + b} : u^{2}

\frac{u ^{2} ( - a + b )}{- a + b}

Cancel the common factor:

- a + b

= u^{2}

Simplify

\frac{x}{- a + b} - \frac{a}{- a + b} : \frac{x - a}{- a + b}

\frac{x}{- a + b} - \frac{a}{- a + b}

Apply rule

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

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

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

For

x^{2} = f (a)

the solutions are

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

u = \frac{x - a}{- a + b}, u = - \frac{x - a}{- a + b}; - a + b \neq = 0

Substitute back

u = sin (t)

sin (t) = \frac{x - a}{- a + b}, sin (t) = - \frac{x - a}{- a + b}; - a + b \neq = 0

sin (t) = \frac{x - a}{- a + b}, sin (t) = - \frac{x - a}{- a + b}; - a + b \neq = 0

sin (t) = \frac{x - a}{- a + b} : t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

sin (t) = \frac{x - a}{- a + b}

Apply trig inverse properties

sin (t) = \frac{x - a}{- a + b}

General solutions for

sin (t) = \frac{x - a}{- a + b}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

sin (t) = - \frac{x - a}{- a + b} : t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

sin (t) = - \frac{x - a}{- a + b}

Apply trig inverse properties

sin (t) = - \frac{x - a}{- a + b}

General solutions for

sin (t) = - \frac{x - a}{- a + b}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

Combine all the solutions

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor t,x=acos^2(t)+bsin^2(t) ?
The general solution for solvefor t,x=acos^2(t)+bsin^2(t) is