What is the general solution for cos(4x)+sin(6x)=0 ?

The general solution for cos(4x)+sin(6x)=0 is x=-(pi+4pin)/4 ,x=-(pi+4pin)/(20)

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cos(4x)+sin(6x)=0

cos (4 x) + sin (6 x) = 0

x = - \frac{π + 4 πn}{4}, x = - \frac{π + 4 πn}{20}

Solution steps

cos (4 x) + sin (6 x) = 0

Subtract

sin (6 x)

from both sides

cos (4 x) = - sin (6 x)

Rewrite using trig identities

sin (\frac{π}{2} - 4 x) = sin (- (6 x))

Apply trig inverse properties

- (6 x) = \frac{π}{2} - 4 x + 2 πn, - (6 x) = π - (\frac{π}{2} - 4 x) + 2 πn

- (6 x) = \frac{π}{2} - 4 x + 2 πn : x = - \frac{π + 4 πn}{4}

- (6 x) = π - (\frac{π}{2} - 4 x) + 2 πn : x = - \frac{π + 4 πn}{20}

x = - \frac{π + 4 πn}{4}, x = - \frac{π + 4 πn}{20}

- 15.44 = - 2 sin (- 3 θ + 300)

sin (x) = - cos (3 4^{\circ})

2 sin^{2} (x) = cos^{3} (x) tan (x)

tan (x - 1 0^{\circ}) cot (2 0^{\circ} - x) = 1

3 csc (2 x) - 4 sin (2 x) = 0

What is the general solution for cos(4x)+sin(6x)=0 ?
The general solution for cos(4x)+sin(6x)=0 is x=-(pi+4pin)/4 ,x=-(pi+4pin)/(20)