What is the general solution for solvefor t,cos(2t)-sin(t)=sqrt(o) ?

The general solution for solvefor t,cos(2t)-sin(t)=sqrt(o) is

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solvefor t,cos(2t)-sin(t)=sqrt(o)

Solution

solve for

t, cos (2 t) - sin (t) = o

Solution

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn, t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

Solution steps

cos (2 t) - sin (t) = o

Subtract

o

from both sides

cos (2 t) - sin (t) - o = 0

Rewrite using trig identities

cos (2 t) - sin (t) - o

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (t) - sin (t) - o

1 - sin (t) - o - 2 sin^{2} (t) = 0

Solve by substitution

1 - sin (t) - o - 2 sin^{2} (t) = 0

Let:

sin (t) = u

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

1 - u - o - 2 u^{2} = 0 : u = - \frac{1 + - 8 o + 9}{4}, u = - \frac{1 - - 8 o + 9}{4}

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 1 - o

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) ( 1 - o )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) ( 1 - o )}{2 ( - 2 )}

Simplify

(- 1)^{2} - 4 (- 2) (1 - o) : - 8 o + 9

(- 1)^{2} - 4 (- 2) (1 - o)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 (1 - o)

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 (1 - o) = 8 (1 - o)

4 \cdot 2 (1 - o)

Multiply the numbers:

4 \cdot 2 = 8

= 8 (- o + 1)

= 1 + 8 (- o + 1)

Expand

1 + 8 (1 - o) : - 8 o + 9

1 + 8 (1 - o)

Expand

8 (1 - o) : 8 - 8 o

8 (1 - o)

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 1, c = o

= 8 \cdot 1 - 8 o

Multiply the numbers:

8 \cdot 1 = 8

= 8 - 8 o

= 1 + 8 - 8 o

Add the numbers:

1 + 8 = 9

= - 8 o + 9

= - 8 o + 9

u_{1, 2} = \frac{- ( - 1 ) \pm - 8 o + 9}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + - 8 o + 9}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - - 8 o + 9}{2 ( - 2 )}

u = \frac{- ( - 1 ) + - 8 o + 9}{2 ( - 2 )} : - \frac{1 + - 8 o + 9}{4}

\frac{- ( - 1 ) + - 8 o + 9}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + - 8 o + 9}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + - 8 o + 9}{- 4}

Apply the fraction rule:

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

= - \frac{1 + - 8 o + 9}{4}

u = \frac{- ( - 1 ) - - 8 o + 9}{2 ( - 2 )} : - \frac{1 - - 8 o + 9}{4}

\frac{- ( - 1 ) - - 8 o + 9}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - - 8 o + 9}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - - 8 o + 9}{- 4}

Apply the fraction rule:

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

= - \frac{1 - - 8 o + 9}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + - 8 o + 9}{4}, u = - \frac{1 - - 8 o + 9}{4}

Substitute back

u = sin (t)

sin (t) = - \frac{1 + - 8 o + 9}{4}, sin (t) = - \frac{1 - - 8 o + 9}{4}

sin (t) = - \frac{1 + - 8 o + 9}{4}, sin (t) = - \frac{1 - - 8 o + 9}{4}

sin (t) = - \frac{1 + - 8 o + 9}{4} : t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn

sin (t) = - \frac{1 + - 8 o + 9}{4}

Apply trig inverse properties

sin (t) = - \frac{1 + - 8 o + 9}{4}

General solutions for

sin (t) = - \frac{1 + - 8 o + 9}{4}

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

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn

sin (t) = - \frac{1 - - 8 o + 9}{4} : t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

sin (t) = - \frac{1 - - 8 o + 9}{4}

Apply trig inverse properties

sin (t) = - \frac{1 - - 8 o + 9}{4}

General solutions for

sin (t) = - \frac{1 - - 8 o + 9}{4}

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

t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

Combine all the solutions

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn, t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

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tan (x^{\circ}) = \frac{30}{73}

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor t,cos(2t)-sin(t)=sqrt(o) ?
The general solution for solvefor t,cos(2t)-sin(t)=sqrt(o) is