What is the general solution for cos(2x-15)=-sin(60-3x) ?

The general solution for cos(2x-15)=-sin(60-3x) is x=(4320n+1980)/(60),x=180+360n-45

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cos(2x-15)=-sin(60-3x)

cos (2 x - 1 5^{\circ}) = - sin (6 0^{\circ} - 3 x)

x = \frac{432 0 ^{\circ} n + 198 0 ^{\circ}}{60}, x = 18 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

Solution steps

cos (2 x - 1 5^{\circ}) = - sin (6 0^{\circ} - 3 x)

Rewrite using trig identities

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

Apply trig inverse properties

- (6 0^{\circ} - 3 x) = 9 0^{\circ} - (2 x - 1 5^{\circ}) + 36 0^{\circ} n, - (6 0^{\circ} - 3 x) = 18 0^{\circ} - (9 0^{\circ} - (2 x - 1 5^{\circ})) + 36 0^{\circ} n

- (6 0^{\circ} - 3 x) = 9 0^{\circ} - (2 x - 1 5^{\circ}) + 36 0^{\circ} n : x = \frac{432 0 ^{\circ} n + 198 0 ^{\circ}}{60}

- (6 0^{\circ} - 3 x) = 18 0^{\circ} - (9 0^{\circ} - (2 x - 1 5^{\circ})) + 36 0^{\circ} n : x = 18 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

x = \frac{432 0 ^{\circ} n + 198 0 ^{\circ}}{60}, x = 18 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

4 sin^{2} (x) - cos (x) - 1 = 0

tan (\frac{θ}{2} - \frac{π}{4}) = 1, 0 \leq θ < 2 π

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

2 sin^{2} (x) + sin (x) - 6 = 0

tan (θ) = \frac{8}{15}, sin (θ) > 0

What is the general solution for cos(2x-15)=-sin(60-3x) ?
The general solution for cos(2x-15)=-sin(60-3x) is x=(4320n+1980)/(60),x=180+360n-45