What is the general solution for 2(cos(t))2-cos(t)-1=0 ?

The general solution for 2(cos(t))2-cos(t)-1=0 is t=1.23095…+2pin,t=2pi-1.23095…+2pin

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2(cos(t))2-cos(t)-1=0

2 (cos (t)) 2 - cos (t) - 1 = 0

t = 1.23095 \dots + 2 πn, t = 2 π - 1.23095 \dots + 2 πn

Degrees

t = 70.52877 \dots^{\circ} + 36 0^{\circ} n, t = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 (cos (t)) \cdot 2 - cos (t) - 1 = 0

Simplify

2 cos (t) \cdot 2 - cos (t) - 1 : 3 cos (t) - 1

2 cos (t) \cdot 2 - cos (t) - 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos (t) - cos (t) - 1

Add similar elements:

4 cos (t) - cos (t) = 3 cos (t)

= 3 cos (t) - 1

3 cos (t) - 1 = 0

Move

1

to the right side

3 cos (t) - 1 = 0

Add

1

to both sides

3 cos (t) - 1 + 1 = 0 + 1

Simplify

3 cos (t) = 1

3 cos (t) = 1

Divide both sides by

3

3 cos (t) = 1

Divide both sides by

3

\frac{3 cos ( t )}{3} = \frac{1}{3}

Simplify

cos (t) = \frac{1}{3}

cos (t) = \frac{1}{3}

Apply trig inverse properties

cos (t) = \frac{1}{3}

General solutions for

cos (t) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

Show solutions in decimal form

t = 1.23095 \dots + 2 πn, t = 2 π - 1.23095 \dots + 2 πn

x, y^{(i v)} - 16 y = - 8 cosh (2 x)

cos (c) = cos (7) cos (6 0^{\circ})

cos (x π) = (- 1)^{n}

3 csc (θ) = 4.5

cos^{3} (x) = \frac{1}{4 ( 3 cos ( x ) + cos ( 3 x ) )}

What is the general solution for 2(cos(t))2-cos(t)-1=0 ?
The general solution for 2(cos(t))2-cos(t)-1=0 is t=1.23095…+2pin,t=2pi-1.23095…+2pin