What is the general solution for 1+cos(2t)=cos^2(t) ?

The general solution for 1+cos(2t)=cos^2(t) is t= pi/2+2pin,t=(3pi)/2+2pin

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

Solution

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

Solution

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

cos^{2} (t)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

1 + cos (2 t) - (cos (2 t) + sin^{2} (t)) : 1 - sin^{2} (t)

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

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

Add similar elements:

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

= 1 - sin^{2} (t)

= 1 - sin^{2} (t)

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

Solve by substitution

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

Let:

sin (t) = u

1 - u^{2} = 0

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sin (t)

sin (t) = 1, sin (t) = - 1

sin (t) = 1, sin (t) = - 1

sin (t) = 1 : t = \frac{π}{2} + 2 πn

sin (t) = 1

General solutions for

sin (t) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

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

sin (t) = - 1

General solutions for

sin (t) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = \frac{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

Popular Examples

2sin(2x+(3pi)/2)+1=0

2 sin (2 x + \frac{3 π}{2}) + 1 = 0

2cos(2x)-sin(x)-1=0

2 cos (2 x) - sin (x) - 1 = 0

2cos^2(θ)-2sin^2(θ)=sqrt(2)

2 cos^{2} (θ) - 2 sin^{2} (θ) = 2

cot(θ)+3csc(θ)=5

cot (θ) + 3 csc (θ) = 5

2cos^2(θ)-2sin^2(θ)=sqrt(3)

2 cos^{2} (θ) - 2 sin^{2} (θ) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 1+cos(2t)=cos^2(t) ?
The general solution for 1+cos(2t)=cos^2(t) is t= pi/2+2pin,t=(3pi)/2+2pin