What is the general solution for sin(270+x)-cos(180-x)=-sin(x) ?

The general solution for sin(270+x)-cos(180-x)=-sin(x) is x=360n,x=180+360n

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sin(270+x)-cos(180-x)=-sin(x)

sin (27 0^{\circ} + x) - cos (18 0^{\circ} - x) = - sin (x)

x = 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (27 0^{\circ} + x) - cos (18 0^{\circ} - x) = - sin (x)

Rewrite using trig identities

0 = - sin (x)

Subtract

- sin (x)

from both sides

sin (x) = 0

General solutions for

sin (x) = 0

x = 0 + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solve

x = 0 + 36 0^{\circ} n : x = 36 0^{\circ} n

x = 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

9 cos (3 x) = 0

16 (l) cos (θ) = 5 sin (2 θ) (2 l)

∣ sin (x) ∣ = \frac{2}{π}

2 + cos (2 θ) = 0

cos (\frac{π ( x + 5 )}{3}) = \frac{1}{2}

What is the general solution for sin(270+x)-cos(180-x)=-sin(x) ?
The general solution for sin(270+x)-cos(180-x)=-sin(x) is x=360n,x=180+360n