What is the general solution for sin(20-θ)-cos(10+3θ)=0 ?

The general solution for sin(20-θ)-cos(10+3θ)=0 is θ=(1080n+180)/6 ,θ=-(180(9n+2))/(18)

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sin(20-θ)-cos(10+3θ)=0

sin (2 0^{\circ} - θ) - cos (1 0^{\circ} + 3 θ) = 0

θ = \frac{108 0 ^{\circ} n + 18 0 ^{\circ}}{6}, θ = - \frac{18 0 ^{\circ} ( 9 n + 2 )}{18}

Solution steps

sin (2 0^{\circ} - θ) - cos (1 0^{\circ} + 3 θ) = 0

Add

cos (1 0^{\circ} + 3 θ)

to both sides

sin (2 0^{\circ} - θ) = cos (1 0^{\circ} + 3 θ)

Rewrite using trig identities

sin (2 0^{\circ} - θ) = sin (9 0^{\circ} - (1 0^{\circ} + 3 θ))

Apply trig inverse properties

2 0^{\circ} - θ = 9 0^{\circ} - (1 0^{\circ} + 3 θ) + 36 0^{\circ} n, 2 0^{\circ} - θ = 18 0^{\circ} - (9 0^{\circ} - (1 0^{\circ} + 3 θ)) + 36 0^{\circ} n

2 0^{\circ} - θ = 9 0^{\circ} - (1 0^{\circ} + 3 θ) + 36 0^{\circ} n : θ = \frac{108 0 ^{\circ} n + 18 0 ^{\circ}}{6}

2 0^{\circ} - θ = 18 0^{\circ} - (9 0^{\circ} - (1 0^{\circ} + 3 θ)) + 36 0^{\circ} n : θ = - \frac{18 0 ^{\circ} ( 9 n + 2 )}{18}

θ = \frac{108 0 ^{\circ} n + 18 0 ^{\circ}}{6}, θ = - \frac{18 0 ^{\circ} ( 9 n + 2 )}{18}

tan (x) = \frac{1}{1}

cos (2 θ) + 3 sin (θ) = 2

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

sec^{2} (\frac{x}{3}) = 2

sin (w) = 0.79

What is the general solution for sin(20-θ)-cos(10+3θ)=0 ?
The general solution for sin(20-θ)-cos(10+3θ)=0 is θ=(1080n+180)/6 ,θ=-(180(9n+2))/(18)