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

The general solution for cos(x-1)= 1/2 ,0<= x<= 360 is x=(180+3)/3 ,x=(900+3)/3

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

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

x = \frac{18 0 ^{\circ} + 3}{3}, x = \frac{90 0 ^{\circ} + 3}{3}

Solution steps

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

General solutions for

cos (x - 1) = \frac{1}{2}

x - 1 = 6 0^{\circ} + 36 0^{\circ} n, x - 1 = 30 0^{\circ} + 36 0^{\circ} n

Solve

x - 1 = 6 0^{\circ} + 36 0^{\circ} n : x = 6 0^{\circ} + 36 0^{\circ} n + 1

Solve

x - 1 = 30 0^{\circ} + 36 0^{\circ} n : x = 30 0^{\circ} + 36 0^{\circ} n + 1

x = 6 0^{\circ} + 36 0^{\circ} n + 1, x = 30 0^{\circ} + 36 0^{\circ} n + 1

Solutions for the range

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

x = \frac{18 0 ^{\circ} + 3}{3}, x = \frac{90 0 ^{\circ} + 3}{3}

cos (t) = \frac{21}{29}

cos (x) = sin (x - \frac{π}{3})

sin (x + \frac{π}{4}) + sin (x + \frac{π}{4}) = - 1

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

4 sin^{2} (x) + 9 = 12

What is the general solution for cos(x-1)= 1/2 ,0<= x<= 360 ?
The general solution for cos(x-1)= 1/2 ,0<= x<= 360 is x=(180+3)/3 ,x=(900+3)/3