What is the general solution for sin(2x)=cos(x-15) ?

The general solution for sin(2x)=cos(x-15) is x=(4320n+1260)/(36),x=180-105+360n

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sin(2x)=cos(x-15)

sin (2 x) = cos (x - 1 5^{\circ})

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}, x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

Solution steps

sin (2 x) = cos (x - 1 5^{\circ})

Rewrite using trig identities

sin (2 x) = sin (9 0^{\circ} - (x - 1 5^{\circ}))

Apply trig inverse properties

2 x = 9 0^{\circ} - (x - 1 5^{\circ}) + 36 0^{\circ} n, 2 x = 18 0^{\circ} - (9 0^{\circ} - (x - 1 5^{\circ})) + 36 0^{\circ} n

2 x = 9 0^{\circ} - (x - 1 5^{\circ}) + 36 0^{\circ} n : x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}

2 x = 18 0^{\circ} - (9 0^{\circ} - (x - 1 5^{\circ})) + 36 0^{\circ} n : x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

x = \frac{432 0 ^{\circ} n + 126 0 ^{\circ}}{36}, x = 18 0^{\circ} - 10 5^{\circ} + 36 0^{\circ} n

(1 + cos (4 x)) sin (4 x) = cos^{2} (2 x)

arcsin (\frac{3}{5}) + arccos (x) = π

cos (θ) = \frac{6}{7}

2 cos (2 θ) + 7 sin (θ) = 5

tan (x^{2}) = 1

What is the general solution for sin(2x)=cos(x-15) ?
The general solution for sin(2x)=cos(x-15) is x=(4320n+1260)/(36),x=180-105+360n