What is the general solution for sin(4x)+6cos(2x)=0 ?

The general solution for sin(4x)+6cos(2x)=0 is x=(pi+4pin)/4 ,x=(3pi+4pin)/4

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sin(4x)+6cos(2x)=0

sin (4 x) + 6 cos (2 x) = 0

x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}

Solution steps

sin (4 x) + 6 cos (2 x) = 0

Let:

u = 2 x

sin (2 u) + 6 cos (u) = 0

Rewrite using trig identities

6 cos (u) + 2 cos (u) sin (u) = 0

Factor

6 cos (u) + 2 cos (u) sin (u) : 2 cos (u) (sin (u) + 3)

2 cos (u) (sin (u) + 3) = 0

Solving each part separately

cos (u) = 0 or sin (u) + 3 = 0

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

sin (u) + 3 = 0 :

No Solution

Combine all the solutions

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

Substitute back

u = 2 x

2 x = \frac{π}{2} + 2 πn : x = \frac{π + 4 πn}{4}

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π + 4 πn}{4}

x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}

sec (θ) + tan (θ) = 0

sin (θ) = \frac{3.5}{5}

sec (a) = \frac{41}{12}, 2 π < a < \frac{5 π}{2}, sin (\frac{a}{2})

sin (2 x) = 0.3

sin (2 x) = 0.6

What is the general solution for sin(4x)+6cos(2x)=0 ?
The general solution for sin(4x)+6cos(2x)=0 is x=(pi+4pin)/4 ,x=(3pi+4pin)/4