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

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

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cos(2x)=1,0<= x<= 2pi

cos (2 x) = 1, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

Solution steps

cos (2 x) = 1, 0 \leq x \leq 2 π

General solutions for

cos (2 x) = 1

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

x = πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

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

2 sin (θ) - 1 = 0

sin (\frac{19 π}{12})

sin (36 0^{\circ})

sin (θ) = - \frac{1}{2}

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