What is the general solution for sin(40-x)=cos(3x) ?

The general solution for sin(40-x)=cos(3x) is x=(6480n+900)/(36),x=-(900+6480n)/(72)

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sin(40-x)=cos(3x)

sin (4 0^{\circ} - x) = cos (3 x)

x = \frac{648 0 ^{\circ} n + 90 0 ^{\circ}}{36}, x = - \frac{90 0 ^{\circ} + 648 0 ^{\circ} n}{72}

Solution steps

sin (4 0^{\circ} - x) = cos (3 x)

Rewrite using trig identities

sin (4 0^{\circ} - x) = sin (9 0^{\circ} - 3 x)

Apply trig inverse properties

4 0^{\circ} - x = 9 0^{\circ} - 3 x + 36 0^{\circ} n, 4 0^{\circ} - x = 18 0^{\circ} - (9 0^{\circ} - 3 x) + 36 0^{\circ} n

4 0^{\circ} - x = 9 0^{\circ} - 3 x + 36 0^{\circ} n : x = \frac{648 0 ^{\circ} n + 90 0 ^{\circ}}{36}

4 0^{\circ} - x = 18 0^{\circ} - (9 0^{\circ} - 3 x) + 36 0^{\circ} n : x = - \frac{90 0 ^{\circ} + 648 0 ^{\circ} n}{72}

x = \frac{648 0 ^{\circ} n + 90 0 ^{\circ}}{36}, x = - \frac{90 0 ^{\circ} + 648 0 ^{\circ} n}{72}

y, e^{x} - sin (y) = x

cot (x) - tan (x) = 23

cos (5 t) cos (3 t) = \frac{1}{2} + sin (- 5 t) sin (3 t)

sin (3 θ + 7 2^{\circ}) = cos (4 8^{\circ}), 0^{\circ} \leq θ \leq 36 0^{\circ}

cot^{2} (y) + csc (y) - 5 = 0

What is the general solution for sin(40-x)=cos(3x) ?
The general solution for sin(40-x)=cos(3x) is x=(6480n+900)/(36),x=-(900+6480n)/(72)