What is the general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi ?

The general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi is x=(25pi)/(28),x=(39pi)/(28)

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cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi

cos (x - \frac{π}{7}) = \frac{- 2}{2}, 0 \leq x \leq 2 π

x = \frac{25 π}{28}, x = \frac{39 π}{28}

Solution steps

cos (x - \frac{π}{7}) = \frac{- 2}{2}, 0 \leq x \leq 2 π

Apply the fraction rule:

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

cos (x - \frac{π}{7}) = - \frac{2}{2}

General solutions for

cos (x - \frac{π}{7}) = - \frac{2}{2}

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn, x - \frac{π}{7} = \frac{5 π}{4} + 2 πn

Solve

x - \frac{π}{7} = \frac{3 π}{4} + 2 πn : x = 2 πn + \frac{25 π}{28}

Solve

x - \frac{π}{7} = \frac{5 π}{4} + 2 πn : x = 2 πn + \frac{39 π}{28}

x = 2 πn + \frac{25 π}{28}, x = 2 πn + \frac{39 π}{28}

Solutions for the range

0 \leq x \leq 2 π

x = \frac{25 π}{28}, x = \frac{39 π}{28}

sin (2 θ) = sin (θ)

25 cos^{2} (θ) - 1 = 0

cos (θ) = \frac{4}{7}

cos (θ) - 1 = - 1

2 = sec (θ)

What is the general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi ?
The general solution for cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pi is x=(25pi)/(28),x=(39pi)/(28)