What is the general solution for 2/pi arccos(m*1-1)=0 ?

The general solution for 2/pi arccos(m*1-1)=0 is m=2

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2/pi arccos(m*1-1)=0

\frac{2}{π} arccos (m \cdot 1 - 1) = 0

m = 2

Solution steps

\frac{2}{π} arccos (m \cdot 1 - 1) = 0

Multiply both sides by

π

\frac{2}{π} arccos (m 1 - 1) = 0

Multiply both sides by

π

\frac{2}{π} arccos (m 1 - 1) π = 0 π

Simplify

2 arccos (m 1 - 1) = 0

2 arccos (m 1 - 1) = 0

Divide both sides by

2

2 arccos (m 1 - 1) = 0

Divide both sides by

2

\frac{2 arccos ( m 1 - 1 )}{2} = \frac{0}{2}

Simplify

arccos (m 1 - 1) = 0

arccos (m 1 - 1) = 0

Apply trig inverse properties

arccos (m \cdot 1 - 1) = 0

arccos (x) = a \Rightarrow x = cos (a)

m \cdot 1 - 1 = cos (0)

cos (0) = 1

cos (0)

Use the following trivial identity:

cos (0) = 1

cos (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}

= 1

= 1

m \cdot 1 - 1 = 1

m \cdot 1 - 1 = 1

Solve

m \cdot 1 - 1 = 1 : m = 2

m \cdot 1 - 1 = 1

Multiply:

m \cdot 1 = m

m - 1 = 1

Move

1

to the right side

m - 1 = 1

Add

1

to both sides

m - 1 + 1 = 1 + 1

Simplify

m = 2

m = 2

m = 2

sin (θ) cos (θ) - \frac{1}{2} cos (θ) = 0

cos^{2} (60) + cos^{2} (45) + cos^{2} (x) = 1

3 sin (5 x) = - 1, x < 2 π, 0

3 cos (x) + cos (3 x) = 0

sin^{2} (a) = 2 sin (a) cos (a)

What is the general solution for 2/pi arccos(m*1-1)=0 ?
The general solution for 2/pi arccos(m*1-1)=0 is m=2