What is the general solution for cos(60-x)=sin(70-3x) ?

The general solution for cos(60-x)=sin(70-3x) is x=-(2160n+180-420)/(24),x=-(900+2160n-420)/(12)

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cos(60-x)=sin(70-3x)

cos (6 0^{\circ} - x) = sin (70 - 3 x)

x = - \frac{216 0 ^{\circ} n + 18 0 ^{\circ} - 420}{24}, x = - \frac{90 0 ^{\circ} + 216 0 ^{\circ} n - 420}{12}

Solution steps

cos (6 0^{\circ} - x) = sin (70 - 3 x)

Rewrite using trig identities

cos (6 0^{\circ} - x) = sin (9 0^{\circ} - (6 0^{\circ} - x))

Apply trig inverse properties

70 - 3 x = 9 0^{\circ} - (6 0^{\circ} - x) + 36 0^{\circ} n, 70 - 3 x = 18 0^{\circ} - (9 0^{\circ} - (6 0^{\circ} - x)) + 36 0^{\circ} n

70 - 3 x = 9 0^{\circ} - (6 0^{\circ} - x) + 36 0^{\circ} n : x = - \frac{216 0 ^{\circ} n + 18 0 ^{\circ} - 420}{24}

70 - 3 x = 18 0^{\circ} - (9 0^{\circ} - (6 0^{\circ} - x)) + 36 0^{\circ} n : x = - \frac{90 0 ^{\circ} + 216 0 ^{\circ} n - 420}{12}

x = - \frac{216 0 ^{\circ} n + 18 0 ^{\circ} - 420}{24}, x = - \frac{90 0 ^{\circ} + 216 0 ^{\circ} n - 420}{12}

sin (- x) = sin (x)

sin (x) = a

sin (6 x) - sin (2 x) = 0

z, cos (x) + cos (y) + cos (z) = 0

12 sin^{2} (x) + cos (x) - 6 = 0

What is the general solution for cos(60-x)=sin(70-3x) ?
The general solution for cos(60-x)=sin(70-3x) is x=-(2160n+180-420)/(24),x=-(900+2160n-420)/(12)