What is the general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) ?

The general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) is x=pin-1/2 ,x= pi/2-1/2+pin

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sin(0.5pi)cos(0.5pi)=0.5sin(2x+1)

sin (0.5 π) cos (0.5 π) = 0.5 sin (2 x + 1)

x = πn - \frac{1}{2}, x = \frac{π}{2} - \frac{1}{2} + πn

Solution steps

sin (0.5 π) cos (0.5 π) = 0.5 sin (2 x + 1)

sin (0.5 π) = 1

cos (0.5 π) = 0

1 \cdot 0 = 0.5 sin (2 x + 1)

Switch sides

0.5 sin (2 x + 1) = 1 \cdot 0

Apply rule

0 \cdot a = 0

0.5 sin (2 x + 1) = 0

Multiply both sides by

10

5 sin (2 x + 1) = 0

Divide both sides by

5

sin (2 x + 1) = 0

General solutions for

sin (2 x + 1) = 0

2 x + 1 = 0 + 2 πn, 2 x + 1 = π + 2 πn

Solve

2 x + 1 = 0 + 2 πn : x = πn - \frac{1}{2}

Solve

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

x = πn - \frac{1}{2}, x = \frac{π}{2} - \frac{1}{2} + πn

2 sin^{2} (u) = 1 + sin (u), 0 \leq y (u) < 2 π

sin (α) = cos^{2} (4 5^{\circ})

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

t, 3 cos (5 t) = 2

2 sin (x) cos (x) + 3 cos (x) = 0

What is the general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) ?
The general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) is x=pin-1/2 ,x= pi/2-1/2+pin