What is the general solution for 2sin(t)+cos(2t)=0 ?

The general solution for 2sin(t)+cos(2t)=0 is t=-0.37473…+2pin,t=pi+0.37473…+2pin

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Popular >

2sin(t)+cos(2t)=0

Solution

2 sin (t) + cos (2 t) = 0

Solution

t = - 0.37473 \dots + 2 πn, t = π + 0.37473 \dots + 2 πn

Degrees

t = - 21.47070 \dots^{\circ} + 36 0^{\circ} n, t = 201.47070 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (t) + cos (2 t) = 0

Rewrite using trig identities

cos (2 t) + 2 sin (t)

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

sin (t) = u

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

1 + 2 u - 2 u^{2} = 0 : u = - \frac{- 1 + 3}{2}, u = \frac{1 + 3}{2}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 (- 2) \cdot 1 = 23

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

u_{1, 2} = \frac{- 2 \pm 2 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 2 + 2 3}{2 ( - 2 )}, u_{2} = \frac{- 2 - 2 3}{2 ( - 2 )}

u = \frac{- 2 + 2 3}{2 ( - 2 )} : - \frac{- 1 + 3}{2}

\frac{- 2 + 2 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 2 3}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 3}{- 4}

Apply the fraction rule:

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

= - \frac{- 2 + 2 3}{4}

Cancel

\frac{- 2 + 2 3}{4} : \frac{3 - 1}{2}

\frac{- 2 + 2 3}{4}

Factor

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

Rewrite as

= - 2 \cdot 1 + 23

Factor out common term

2

= 2 (- 1 + 3)

= \frac{2 ( - 1 + 3 )}{4}

Cancel the common factor:

2

= \frac{- 1 + 3}{2}

= - \frac{3 - 1}{2}

= - \frac{- 1 + 3}{2}

u = \frac{- 2 - 2 3}{2 ( - 2 )} : \frac{1 + 3}{2}

\frac{- 2 - 2 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 2 3}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 3}{- 4}

Apply the fraction rule:

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

- 2 - 23 = - (2 + 23)

= \frac{2 + 2 3}{4}

Factor

2 + 23 : 2 (1 + 3)

2 + 23

Rewrite as

= 2 \cdot 1 + 23

Factor out common term

2

= 2 (1 + 3)

= \frac{2 ( 1 + 3 )}{4}

Cancel the common factor:

2

= \frac{1 + 3}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 3}{2}, u = \frac{1 + 3}{2}

Substitute back

u = sin (t)

sin (t) = - \frac{- 1 + 3}{2}, sin (t) = \frac{1 + 3}{2}

sin (t) = - \frac{- 1 + 3}{2}, sin (t) = \frac{1 + 3}{2}

sin (t) = - \frac{- 1 + 3}{2} : t = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, t = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

sin (t) = - \frac{- 1 + 3}{2}

Apply trig inverse properties

sin (t) = - \frac{- 1 + 3}{2}

General solutions for

sin (t) = - \frac{- 1 + 3}{2}

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

t = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, t = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

t = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, t = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

sin (t) = \frac{1 + 3}{2} :

No Solution

sin (t) = \frac{1 + 3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

t = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, t = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

Show solutions in decimal form

t = - 0.37473 \dots + 2 πn, t = π + 0.37473 \dots + 2 πn

Popular Examples

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23 = csc (x) cot (x)

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4 sin^{2} (x) - 4 sin (x) = - 1

sin^2(θ)=3cos^2(θ)

sin^{2} (θ) = 3 cos^{2} (θ)

20sin^2(x)+27cos(x)=6

20 sin^{2} (x) + 27 cos (x) = 6

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(t)+cos(2t)=0 ?
The general solution for 2sin(t)+cos(2t)=0 is t=-0.37473…+2pin,t=pi+0.37473…+2pin