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

The general solution for sin(x)-1=0,0<= x<= 2pi is x= pi/2

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sin(x)-1=0,0<= x<= 2pi

sin (x) - 1 = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}

Solution steps

sin (x) - 1 = 0, 0 \leq x \leq 2 π

Move

1

to the right side

sin (x) = 1

General solutions for

sin (x) = 1

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}

3 tan^{3} (θ) + 3 tan^{2} (θ) - tan (θ) - 1 = 0

sin (4 x) + 2 sin (2 x) = 0

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

2 sec^{2} (x) + 5 sec (x) + 2 = 0

sin^{2} (θ) = 8 (cos (- θ) + 1)

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