What is the general solution for cos(sqrt(2)T)=1,0<= T<= 2pi ?

The general solution for cos(sqrt(2)T)=1,0<= T<= 2pi is T=0,T=sqrt(2)pi

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cos(sqrt(2)T)=1,0<= T<= 2pi

cos (2 T) = 1, 0 \leq T \leq 2 π

T = 0, T = 2 π

Degrees

T = 0^{\circ}, T = 254.55844 \dots^{\circ}

Solution steps

cos (2 T) = 1, 0 \leq T \leq 2 π

General solutions for

cos (2 T) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 T = 0 + 2 πn

2 T = 0 + 2 πn

Solve

2 T = 0 + 2 πn : T = 2 πn

2 T = 0 + 2 πn

0 + 2 πn = 2 πn

2 T = 2 πn

Divide both sides by

2

2 T = 2 πn

Divide both sides by

2

\frac{2 T}{2} = \frac{2 πn}{2}

Simplify

\frac{2 T}{2} = \frac{2 πn}{2}

Simplify

\frac{2 T}{2} : T

\frac{2 T}{2}

Cancel the common factor:

2

= T

Simplify

\frac{2 πn}{2} : 2 πn

\frac{2 πn}{2}

Apply radical rule:

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

= \frac{2 πn}{2 ^{\frac{1}{2}}}

Apply exponent rule:

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

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

= 2^{- \frac{1}{2} + 1} πn

Subtract the numbers:

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

= 2^{\frac{1}{2}} πn

Apply radical rule:

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

= 2 πn

T = 2 πn

T = 2 πn

T = 2 πn

T = 2 πn

Solutions for the range

0 \leq T \leq 2 π

T = 0, T = 2 π

2 - 2 cos^{2} (\frac{x}{2}) = 2 cos^{2} (x)

sec^{2} (θ) - 2 = tan^{2} (θ)

cos (x) = sin (x) + 1

2 sin^{2} (θ) = 3 (1 - cos (- θ))

sec (x) - 2 = 0

What is the general solution for cos(sqrt(2)T)=1,0<= T<= 2pi ?
The general solution for cos(sqrt(2)T)=1,0<= T<= 2pi is T=0,T=sqrt(2)pi