What is the general solution for cos(3x+20)=-sin(x-60) ?

The general solution for cos(3x+20)=-sin(x-60) is x=(2160n+180-120)/(12),x=-(180+2160n+120)/(24)

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cos(3x+20)=-sin(x-60)

cos (3 x + 20) = - sin (x - 6 0^{\circ})

x = \frac{216 0 ^{\circ} n + 18 0 ^{\circ} - 120}{12}, x = - \frac{18 0 ^{\circ} + 216 0 ^{\circ} n + 120}{24}

Solution steps

cos (3 x + 20) = - sin (x - 6 0^{\circ})

Rewrite using trig identities

cos (3 x + 20) = sin (9 0^{\circ} - (3 x + 20))

Apply trig inverse properties

- (x - 6 0^{\circ}) = 9 0^{\circ} - (3 x + 20) + 36 0^{\circ} n, - (x - 6 0^{\circ}) = 18 0^{\circ} - (9 0^{\circ} - (3 x + 20)) + 36 0^{\circ} n

- (x - 6 0^{\circ}) = 9 0^{\circ} - (3 x + 20) + 36 0^{\circ} n : x = \frac{216 0 ^{\circ} n + 18 0 ^{\circ} - 120}{12}

- (x - 6 0^{\circ}) = 18 0^{\circ} - (9 0^{\circ} - (3 x + 20)) + 36 0^{\circ} n : x = - \frac{18 0 ^{\circ} + 216 0 ^{\circ} n + 120}{24}

x = \frac{216 0 ^{\circ} n + 18 0 ^{\circ} - 120}{12}, x = - \frac{18 0 ^{\circ} + 216 0 ^{\circ} n + 120}{24}

cos (2 x) - 3 cos (x) - 4 = 0

3 - 5 cos (θ) = 0

tan (x) = \frac{48}{50}

\frac{sin ( 5 4 ^{\circ} )}{7} = \frac{sin ( x )}{10}

sin (y) = \frac{- 1}{2}

What is the general solution for cos(3x+20)=-sin(x-60) ?
The general solution for cos(3x+20)=-sin(x-60) is x=(2160n+180-120)/(12),x=-(180+2160n+120)/(24)