What is the general solution for sin(3x+48)=cos(10-x) ?

The general solution for sin(3x+48)=cos(10-x) is x=(1620n+360-216)/9 ,x=(900+3240n-432)/(36)

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sin(3x+48)=cos(10-x)

sin (3 x + 48) = cos (1 0^{\circ} - x)

x = \frac{162 0 ^{\circ} n + 36 0 ^{\circ} - 216}{9}, x = \frac{90 0 ^{\circ} + 324 0 ^{\circ} n - 432}{36}

Solution steps

sin (3 x + 48) = cos (1 0^{\circ} - x)

Rewrite using trig identities

sin (3 x + 48) = sin (9 0^{\circ} - (1 0^{\circ} - x))

Apply trig inverse properties

3 x + 48 = 9 0^{\circ} - (1 0^{\circ} - x) + 36 0^{\circ} n, 3 x + 48 = 18 0^{\circ} - (9 0^{\circ} - (1 0^{\circ} - x)) + 36 0^{\circ} n

3 x + 48 = 9 0^{\circ} - (1 0^{\circ} - x) + 36 0^{\circ} n : x = \frac{162 0 ^{\circ} n + 36 0 ^{\circ} - 216}{9}

3 x + 48 = 18 0^{\circ} - (9 0^{\circ} - (1 0^{\circ} - x)) + 36 0^{\circ} n : x = \frac{90 0 ^{\circ} + 324 0 ^{\circ} n - 432}{36}

x = \frac{162 0 ^{\circ} n + 36 0 ^{\circ} - 216}{9}, x = \frac{90 0 ^{\circ} + 324 0 ^{\circ} n - 432}{36}

0 = sin (5 x)

2 sin (\frac{θ}{3}) + 3 = 0, 0 \leq θ \leq 2 π

sin (2 x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

2 sin (x - \frac{π}{3}) + 3 = 0

sin (x) = 0.77

What is the general solution for sin(3x+48)=cos(10-x) ?
The general solution for sin(3x+48)=cos(10-x) is x=(1620n+360-216)/9 ,x=(900+3240n-432)/(36)