What is the general solution for 6cos(2x)+8sin(2x)=0 ?

The general solution for 6cos(2x)+8sin(2x)=0 is x=-(0.64350…)/2+(pin)/2

Popular >

6cos(2x)+8sin(2x)=0

6 cos (2 x) + 8 sin (2 x) = 0

x = - \frac{0.64350 \dots}{2} + \frac{πn}{2}

Solution steps

6 cos (2 x) + 8 sin (2 x) = 0

Rewrite using trig identities

6 + 8 tan (2 x) = 0

Move

6

to the right side

8 tan (2 x) = - 6

Divide both sides by

8

tan (2 x) = - \frac{3}{4}

Apply trig inverse properties

2 x = arctan (- \frac{3}{4}) + πn

Solve

2 x = arctan (- \frac{3}{4}) + πn : x = - \frac{arctan ( \frac{3}{4} )}{2} + \frac{πn}{2}

x = - \frac{arctan ( \frac{3}{4} )}{2} + \frac{πn}{2}

Show solutions in decimal form

x = - \frac{0.64350 \dots}{2} + \frac{πn}{2}

cos (x) = \frac{8}{17}, sin (x)

\frac{50}{sin ( 10 5 ^{\circ} )} = \frac{20}{sin ( x )}

x, sin^{2} (x) = \frac{5 m - 5}{2}

csc (θ) = - 1.343

csc (θ) = - \frac{3}{2}, π < θ < \frac{3 π}{2}, sin (θ)

What is the general solution for 6cos(2x)+8sin(2x)=0 ?
The general solution for 6cos(2x)+8sin(2x)=0 is x=-(0.64350…)/2+(pin)/2