What is the general solution for 5sin(2x)+3cos(2x)=0 ?

The general solution for 5sin(2x)+3cos(2x)=0 is x=-(0.54041…)/2+(pin)/2

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5sin(2x)+3cos(2x)=0

5 sin (2 x) + 3 cos (2 x) = 0

x = - \frac{0.54041 \dots}{2} + \frac{πn}{2}

Solution steps

5 sin (2 x) + 3 cos (2 x) = 0

Rewrite using trig identities

5 tan (2 x) + 3 = 0

Move

3

to the right side

5 tan (2 x) = - 3

Divide both sides by

5

tan (2 x) = - \frac{3}{5}

Apply trig inverse properties

2 x = arctan (- \frac{3}{5}) + πn

Solve

2 x = arctan (- \frac{3}{5}) + πn : x = - \frac{arctan ( \frac{3}{5} )}{2} + \frac{πn}{2}

x = - \frac{arctan ( \frac{3}{5} )}{2} + \frac{πn}{2}

Show solutions in decimal form

x = - \frac{0.54041 \dots}{2} + \frac{πn}{2}

x, sin (2 x) - sin (x) = 0

x, sin (x^{e}) + cos (y^{e}) = 1

sin^{2} (x) + 3 cos (x) - 1 = 2

cot^{2} (θ) - 3 csc (θ) + 3 = 0

x, arctan (x) + arctan (y) = \frac{π}{2}

What is the general solution for 5sin(2x)+3cos(2x)=0 ?
The general solution for 5sin(2x)+3cos(2x)=0 is x=-(0.54041…)/2+(pin)/2