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

The general solution for 2cos(2t)+2cos^2(t)=0 is t=-0.95531…+pin,t=0.95531…+pin

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Popular >

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

Solution

2 cos (2 t) + 2 cos^{2} (t) = 0

Solution

t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

Degrees

t = - 54.73561 \dots^{\circ} + 18 0^{\circ} n, t = 54.73561 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (2 t) + 2 cos^{2} (t) = 0

Rewrite using trig identities

2 cos (2 t) + 2 cos^{2} (t)

Use the Double Angle identity:

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

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

Simplify

2 (cos^{2} (t) - sin^{2} (t)) + 2 cos^{2} (t) : 4 cos^{2} (t) - 2 sin^{2} (t)

2 (cos^{2} (t) - sin^{2} (t)) + 2 cos^{2} (t)

Expand

2 (cos^{2} (t) - sin^{2} (t)) : 2 cos^{2} (t) - 2 sin^{2} (t)

2 (cos^{2} (t) - sin^{2} (t))

Apply the distributive law:

a (b - c) = ab - a c

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

= 2 cos^{2} (t) - 2 sin^{2} (t)

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

Add similar elements:

2 cos^{2} (t) + 2 cos^{2} (t) = 4 cos^{2} (t)

= 4 cos^{2} (t) - 2 sin^{2} (t)

= 4 cos^{2} (t) - 2 sin^{2} (t)

- 2 sin^{2} (t) + 4 cos^{2} (t) = 0

Factor

- 2 sin^{2} (t) + 4 cos^{2} (t) : 2 (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

- 2 sin^{2} (t) + 4 cos^{2} (t)

Rewrite

4

2 \cdot 2

= - 2 sin^{2} (t) + 2 \cdot 2 cos^{2} (t)

Factor out common term

2

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

Factor

2 cos^{2} (t) - sin^{2} (t) : (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

2 cos^{2} (t) - sin^{2} (t)

Rewrite

2 cos^{2} (t) - sin^{2} (t)

(2 cos (t))^{2} - sin^{2} (t)

2 cos^{2} (t) - sin^{2} (t)

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} cos^{2} (t) - sin^{2} (t)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(2)^{2} cos^{2} (t) = (2 cos (t))^{2}

= (2 cos (t))^{2} - sin^{2} (t)

= (2 cos (t))^{2} - sin^{2} (t)

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

2 cos (t) + sin (t) = 0 : t = arctan (- 2) + πn

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

Rewrite using trig identities

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

Divide both sides by

cos (t), cos (t) \neq = 0

\frac{2 cos ( t ) + sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

Simplify

2 + \frac{sin ( t )}{cos ( t )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 + tan (t) = 0

2 + tan (t) = 0

Move

2

to the right side

2 + tan (t) = 0

Subtract

2

from both sides

2 + tan (t) - 2 = 0 - 2

Simplify

tan (t) = - 2

tan (t) = - 2

Apply trig inverse properties

tan (t) = - 2

General solutions for

tan (t) = - 2

tan (x) = - a \Rightarrow x = arctan (- a) + πn

t = arctan (- 2) + πn

t = arctan (- 2) + πn

2 cos (t) - sin (t) = 0 : t = arctan (2) + πn

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

Rewrite using trig identities

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

Divide both sides by

cos (t), cos (t) \neq = 0

\frac{2 cos ( t ) - sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

Simplify

2 - \frac{sin ( t )}{cos ( t )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 - tan (t) = 0

2 - tan (t) = 0

Move

2

to the right side

2 - tan (t) = 0

Subtract

2

from both sides

2 - tan (t) - 2 = 0 - 2

Simplify

- tan (t) = - 2

- tan (t) = - 2

Divide both sides by

- 1

- tan (t) = - 2

Divide both sides by

- 1

\frac{- tan ( t )}{- 1} = \frac{- 2}{- 1}

Simplify

tan (t) = 2

tan (t) = 2

Apply trig inverse properties

tan (t) = 2

General solutions for

tan (t) = 2

tan (x) = a \Rightarrow x = arctan (a) + πn

t = arctan (2) + πn

t = arctan (2) + πn

Combine all the solutions

t = arctan (- 2) + πn, t = arctan (2) + πn

Show solutions in decimal form

t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2t)+2cos^2(t)=0 ?
The general solution for 2cos(2t)+2cos^2(t)=0 is t=-0.95531…+pin,t=0.95531…+pin