What is the general solution for -1.9sin(t)=0.4cos(2t) ?

The general solution for -1.9sin(t)=0.4cos(2t) is t=-0.19583…+2pin,t=pi+0.19583…+2pin

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Popular >

-1.9sin(t)=0.4cos(2t)

Solution

- 1.9 sin (t) = 0.4 cos (2 t)

Solution

t = - 0.19583 \dots + 2 πn, t = π + 0.19583 \dots + 2 πn

Degrees

t = - 11.22042 \dots^{\circ} + 36 0^{\circ} n, t = 191.22042 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 1.9 sin (t) = 0.4 cos (2 t)

Subtract

0.4 cos (2 t)

from both sides

- 1.9 sin (t) - 0.4 cos (2 t) = 0

Rewrite using trig identities

- 0.4 cos (2 t) - 1.9 sin (t)

Use the Double Angle identity:

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

= - 0.4 (1 - 2 sin^{2} (t)) - 1.9 sin (t)

- (1 - 2 sin^{2} (t)) \cdot 0.4 - 1.9 sin (t) = 0

Solve by substitution

- (1 - 2 sin^{2} (t)) \cdot 0.4 - 1.9 sin (t) = 0

Let:

sin (t) = u

- (1 - 2 u^{2}) \cdot 0.4 - 1.9 u = 0

- (1 - 2 u^{2}) \cdot 0.4 - 1.9 u = 0 : u = \frac{19 + 489}{16}, u = \frac{19 - 489}{16}

- (1 - 2 u^{2}) \cdot 0.4 - 1.9 u = 0

Multiply both sides by

10

- (1 - 2 u^{2}) \cdot 0.4 - 1.9 u = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

- (1 - 2 u^{2}) \cdot 0.4 \cdot 10 - 1.9 u \cdot 10 = 0 \cdot 10

Refine

- 4 (1 - 2 u^{2}) - 19 u = 0

- 4 (1 - 2 u^{2}) - 19 u = 0

Expand

- 4 (1 - 2 u^{2}) - 19 u : - 4 + 8 u^{2} - 19 u

- 4 (1 - 2 u^{2}) - 19 u

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = 2 u^{2}

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

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 \cdot 2 u^{2}

Simplify

- 4 \cdot 1 + 4 \cdot 2 u^{2} : - 4 + 8 u^{2}

- 4 \cdot 1 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 2 = 8

= - 4 + 8 u^{2}

= - 4 + 8 u^{2}

= - 4 + 8 u^{2} - 19 u

- 4 + 8 u^{2} - 19 u = 0

Write in the standard form

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

8 u^{2} - 19 u - 4 = 0

Solve with the quadratic formula

8 u^{2} - 19 u - 4 = 0

Quadratic Equation Formula:

For

a = 8, b = - 19, c = - 4

u_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

(- 19)^{2} - 4 \cdot 8 (- 4) = 489

(- 19)^{2} - 4 \cdot 8 (- 4)

Apply rule

- (- a) = a

= (- 19)^{2} + 4 \cdot 8 \cdot 4

Apply exponent rule:

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

n

is even

(- 19)^{2} = 1 9^{2}

= 1 9^{2} + 4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 1 9^{2} + 128

1 9^{2} = 361

= 361 + 128

Add the numbers:

361 + 128 = 489

= 489

u_{1, 2} = \frac{- ( - 19 ) \pm 489}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 19 ) + 489}{2 \cdot 8}, u_{2} = \frac{- ( - 19 ) - 489}{2 \cdot 8}

u = \frac{- ( - 19 ) + 489}{2 \cdot 8} : \frac{19 + 489}{16}

\frac{- ( - 19 ) + 489}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{19 + 489}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{19 + 489}{16}

u = \frac{- ( - 19 ) - 489}{2 \cdot 8} : \frac{19 - 489}{16}

\frac{- ( - 19 ) - 489}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{19 - 489}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{19 - 489}{16}

The solutions to the quadratic equation are:

u = \frac{19 + 489}{16}, u = \frac{19 - 489}{16}

Substitute back

u = sin (t)

sin (t) = \frac{19 + 489}{16}, sin (t) = \frac{19 - 489}{16}

sin (t) = \frac{19 + 489}{16}, sin (t) = \frac{19 - 489}{16}

sin (t) = \frac{19 + 489}{16} :

No Solution

sin (t) = \frac{19 + 489}{16}

- 1 \leq sin (x) \leq 1

No Solution

sin (t) = \frac{19 - 489}{16} : t = arcsin (\frac{19 - 489}{16}) + 2 πn, t = π + arcsin (- \frac{19 - 489}{16}) + 2 πn

sin (t) = \frac{19 - 489}{16}

Apply trig inverse properties

sin (t) = \frac{19 - 489}{16}

General solutions for

sin (t) = \frac{19 - 489}{16}

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

t = arcsin (\frac{19 - 489}{16}) + 2 πn, t = π + arcsin (- \frac{19 - 489}{16}) + 2 πn

t = arcsin (\frac{19 - 489}{16}) + 2 πn, t = π + arcsin (- \frac{19 - 489}{16}) + 2 πn

Combine all the solutions

t = arcsin (\frac{19 - 489}{16}) + 2 πn, t = π + arcsin (- \frac{19 - 489}{16}) + 2 πn

Show solutions in decimal form

t = - 0.19583 \dots + 2 πn, t = π + 0.19583 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -1.9sin(t)=0.4cos(2t) ?
The general solution for -1.9sin(t)=0.4cos(2t) is t=-0.19583…+2pin,t=pi+0.19583…+2pin