What is the general solution for sin(x+20)=cos(2x+25) ?

The general solution for sin(x+20)=cos(2x+25) is x=(1440n+180)/(12),x=-(3420+12960n)/(36)

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sin(x+20)=cos(2x+25)

sin (x + 2 0^{\circ}) = cos (2 x + 2 5^{\circ})

x = \frac{144 0 ^{\circ} n + 18 0 ^{\circ}}{12}, x = - \frac{342 0 ^{\circ} + 1296 0 ^{\circ} n}{36}

Solution steps

sin (x + 2 0^{\circ}) = cos (2 x + 2 5^{\circ})

Rewrite using trig identities

sin (x + 2 0^{\circ}) = sin (9 0^{\circ} - (2 x + 2 5^{\circ}))

Apply trig inverse properties

x + 2 0^{\circ} = 9 0^{\circ} - (2 x + 2 5^{\circ}) + 36 0^{\circ} n, x + 2 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - (2 x + 2 5^{\circ})) + 36 0^{\circ} n

x + 2 0^{\circ} = 9 0^{\circ} - (2 x + 2 5^{\circ}) + 36 0^{\circ} n : x = \frac{144 0 ^{\circ} n + 18 0 ^{\circ}}{12}

x + 2 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - (2 x + 2 5^{\circ})) + 36 0^{\circ} n : x = - \frac{342 0 ^{\circ} + 1296 0 ^{\circ} n}{36}

x = \frac{144 0 ^{\circ} n + 18 0 ^{\circ}}{12}, x = - \frac{342 0 ^{\circ} + 1296 0 ^{\circ} n}{36}

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

cos (a) = \frac{1}{2}

sin (2 θ) - cos (2 θ) = 0

cos^{2} (x) + sin (x) = 2

4 csc (θ) - 5 = 0

What is the general solution for sin(x+20)=cos(2x+25) ?
The general solution for sin(x+20)=cos(2x+25) is x=(1440n+180)/(12),x=-(3420+12960n)/(36)