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

The general solution for 2cos^2(x-1)=0 is x= pi/2+2pin+1,x=(3pi)/2+2pin+1

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2cos^2(x-1)=0

2 cos^{2} (x - 1) = 0

x = \frac{π}{2} + 2 πn + 1, x = \frac{3 π}{2} + 2 πn + 1

Degrees

x = 147.29577 \dots^{\circ} + 36 0^{\circ} n, x = 327.29577 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x - 1) = 0

Divide both sides by

2

2 cos^{2} (x - 1) = 0

Divide both sides by

2

2 cos^{2} (x - 1) = 0

Divide both sides by

2

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

Simplify

cos^{2} (x - 1) = 0

cos^{2} (x - 1) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x - 1) = 0

General solutions for

cos (x - 1) = 0

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}

x - 1 = \frac{π}{2} + 2 πn, x - 1 = \frac{3 π}{2} + 2 πn

x - 1 = \frac{π}{2} + 2 πn, x - 1 = \frac{3 π}{2} + 2 πn

Solve

x - 1 = \frac{π}{2} + 2 πn : x = \frac{π}{2} + 2 πn + 1

x - 1 = \frac{π}{2} + 2 πn

Move

1

to the right side

x - 1 = \frac{π}{2} + 2 πn

Add

1

to both sides

x - 1 + 1 = \frac{π}{2} + 2 πn + 1

Simplify

x = \frac{π}{2} + 2 πn + 1

x = \frac{π}{2} + 2 πn + 1

Solve

x - 1 = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{2} + 2 πn + 1

x - 1 = \frac{3 π}{2} + 2 πn

Move

1

to the right side

x - 1 = \frac{3 π}{2} + 2 πn

Add

1

to both sides

x - 1 + 1 = \frac{3 π}{2} + 2 πn + 1

Simplify

x = \frac{3 π}{2} + 2 πn + 1

x = \frac{3 π}{2} + 2 πn + 1

x = \frac{π}{2} + 2 πn + 1, x = \frac{3 π}{2} + 2 πn + 1

\frac{1}{cot ^{2} ( x )} + \frac{3}{cot ( x )} = 0

b, \frac{sin ( 6 5 ^{\circ} )}{14} = \frac{sin ( b )}{9}

4 sin (θ) = sin (2 θ)

2 sin (3 x + \frac{3 π}{2}) = 1

n = (2 tan^{3} (2 θ) - 1)^{\frac{1}{2}}

What is the general solution for 2cos^2(x-1)=0 ?
The general solution for 2cos^2(x-1)=0 is x= pi/2+2pin+1,x=(3pi)/2+2pin+1