What is the general solution for sin(2a+10)=cos(3a-20) ?

The general solution for sin(2a+10)=cos(3a-20) is a=(4pin+20+pi}{10},a=-\frac{pi+4pin-60)/2

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sin(2a+10)=cos(3a-20)

sin (2 a + 10) = cos (3 a - 20)

a = \frac{4 πn + 20 + π}{10}, a = - \frac{π + 4 πn - 60}{2}

Solution steps

sin (2 a + 10) = cos (3 a - 20)

Rewrite using trig identities

sin (2 a + 10) = sin (\frac{π}{2} - (3 a - 20))

Apply trig inverse properties

2 a + 10 = \frac{π}{2} - (3 a - 20) + 2 πn, 2 a + 10 = π - (\frac{π}{2} - (3 a - 20)) + 2 πn

2 a + 10 = \frac{π}{2} - (3 a - 20) + 2 πn : a = \frac{4 πn + 20 + π}{10}

2 a + 10 = π - (\frac{π}{2} - (3 a - 20)) + 2 πn : a = - \frac{π + 4 πn - 60}{2}

a = \frac{4 πn + 20 + π}{10}, a = - \frac{π + 4 πn - 60}{2}

b = \frac{3}{( cot ( x ) )}

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

(cot^{2} (x)) - csc (x) = 1

x, u = arctan (\frac{x}{y})

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

What is the general solution for sin(2a+10)=cos(3a-20) ?
The general solution for sin(2a+10)=cos(3a-20) is a=(4pin+20+pi}{10},a=-\frac{pi+4pin-60)/2