What is the general solution for 5sec^2(x)-5=0,0<= x<2pi ?

The general solution for 5sec^2(x)-5=0,0<= x<2pi is x=0,x=pi

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5sec^2(x)-5=0,0<= x<2pi

5 sec^{2} (x) - 5 = 0, 0 \leq x < 2 π

x = 0, x = π

Solution steps

5 sec^{2} (x) - 5 = 0, 0 \leq x < 2 π

Solve by substitution

sec (x) = 1, sec (x) = - 1

sec (x) = 1, 0 \leq x < 2 π : x = 0

sec (x) = - 1, 0 \leq x < 2 π : x = π

Combine all the solutions

x = 0, x = π

2 sin (5 x) + 3 = 2

5 sin (y) - 5 = 3 sin (y) - 4

2 cos^{2} (x) + 3 cos (x) - 3 = 0

3 sin (x) - 5 cos (x) = 0

sin^{2} (x) = 0.85

What is the general solution for 5sec^2(x)-5=0,0<= x<2pi ?
The general solution for 5sec^2(x)-5=0,0<= x<2pi is x=0,x=pi