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

The general solution for cos(pix)=0,0<= x<= 2pi is x= 1/2 ,x= 3/2 ,x= 5/2 ,x= 7/2 ,x= 9/2 ,x= 11/2

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

cos (π x) = 0, 0 \leq x \leq 2 π

x = \frac{1}{2}, x = \frac{3}{2}, x = \frac{5}{2}, x = \frac{7}{2}, x = \frac{9}{2}, x = \frac{11}{2}

Solution steps

cos (π x) = 0, 0 \leq x \leq 2 π

General solutions for

cos (π x) = 0

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

Solve

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

Solve

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

x = \frac{1}{2} + 2 n, x = \frac{3}{2} + 2 n

Solutions for the range

0 \leq x \leq 2 π

x = \frac{1}{2}, x = \frac{3}{2}, x = \frac{5}{2}, x = \frac{7}{2}, x = \frac{9}{2}, x = \frac{11}{2}

cos (2 x - 1 5^{\circ}) = - sin (6 0^{\circ} - 3 x)

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

tan (\frac{θ}{2} - \frac{π}{4}) = 1, 0 \leq θ < 2 π

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

2 sin^{2} (x) + sin (x) - 6 = 0

What is the general solution for cos(pix)=0,0<= x<= 2pi ?
The general solution for cos(pix)=0,0<= x<= 2pi is x= 1/2 ,x= 3/2 ,x= 5/2 ,x= 7/2 ,x= 9/2 ,x= 11/2