What is the general solution for sin(pi/2-x)=-1,0<= x<= 360 ?

The general solution for sin(pi/2-x)=-1,0<= x<= 360 is x=180

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sin(pi/2-x)=-1,0<= x<= 360

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 18 0^{\circ}

Solution steps

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

General solutions for

sin (9 0^{\circ} - x) = - 1

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n

Solve

9 0^{\circ} - x = 27 0^{\circ} + 36 0^{\circ} n : x = - 36 0^{\circ} n - 18 0^{\circ}

x = - 36 0^{\circ} n - 18 0^{\circ}

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 18 0^{\circ}

2 sin (5 x) = - 1

4 sin^{2} (x) + 8 cos (x) - 7 = 0

6 (cos (b))^{2} + cos (b) - 1 = 0

sin (θ) = \frac{1.33}{2.42}

2 tan (x) = 3 - tan^{2} (x), - 2 π \leq, x \leq π

What is the general solution for sin(pi/2-x)=-1,0<= x<= 360 ?
The general solution for sin(pi/2-x)=-1,0<= x<= 360 is x=180