What is the general solution for sin(x-15)=cos(20+x) ?

The general solution for sin(x-15)=cos(20+x) is x=(12960n+3060)/(72)

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sin(x-15)=cos(20+x)

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

x = \frac{1296 0 ^{\circ} n + 306 0 ^{\circ}}{72}

Solution steps

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

Rewrite using trig identities

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

Apply trig inverse properties

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

x - 1 5^{\circ} = 9 0^{\circ} - (2 0^{\circ} + x) + 36 0^{\circ} n : x = \frac{1296 0 ^{\circ} n + 306 0 ^{\circ}}{72}

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

True for all

x; 0 = 18 0^{\circ} + 36 0^{\circ} n - 5 5^{\circ}

Since the equation is undefined for:

True for all

x

x = \frac{1296 0 ^{\circ} n + 306 0 ^{\circ}}{72}

15 sin^{2} (x) - 17 sin (x) + 4 = 0

cos (x) = \frac{451.1}{497}

tan^{2} (x) = 2 sec (x)

- sin^{2} (x) = sin^{4} (x)

3 cos^{2} (x) = sin (2 x) \cdot sin (x)

What is the general solution for sin(x-15)=cos(20+x) ?
The general solution for sin(x-15)=cos(20+x) is x=(12960n+3060)/(72)