What is the general solution for 5sin(2x)+3cos(2x)=0,0<= x<= 180 ?

The general solution for 5sin(2x)+3cos(2x)=0,0<= x<= 180 is x=(-0.54041…+180)/2 ,x=(-0.54041…+360)/2

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5sin(2x)+3cos(2x)=0,0<= x<= 180

5 sin (2 x) + 3 cos (2 x) = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

x = \frac{- 0.54041 \dots + 18 0 ^{\circ}}{2}, x = \frac{- 0.54041 \dots + 36 0 ^{\circ}}{2}

Solution steps

5 sin (2 x) + 3 cos (2 x) = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

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}) + 18 0^{\circ} n

Solve

2 x = arctan (- \frac{3}{5}) + 18 0^{\circ} n : x = - \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

x = - \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = \frac{- arctan ( \frac{3}{5} ) + 18 0 ^{\circ}}{2}, x = \frac{- arctan ( \frac{3}{5} ) + 36 0 ^{\circ}}{2}

Show solutions in decimal form

x = \frac{- 0.54041 \dots + 18 0 ^{\circ}}{2}, x = \frac{- 0.54041 \dots + 36 0 ^{\circ}}{2}

cos (θ) = 0.6798, 0^{\circ} \leq θ \leq 36 0^{\circ}

\frac{2 sin ( x ) cos ( x )}{cos ( x )} = 1

csc (3 θ) = sec (1 5^{\circ})

2 cot^{2} (x) + tan (x) = cot (x)

sin (2 x) = sin (\frac{π}{2} - x)

What is the general solution for 5sin(2x)+3cos(2x)=0,0<= x<= 180 ?
The general solution for 5sin(2x)+3cos(2x)=0,0<= x<= 180 is x=(-0.54041…+180)/2 ,x=(-0.54041…+360)/2