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

Q: 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

Solution

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

Solution

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

Radians

x = \frac{- 0.54041 \dots + π}{2}, x = \frac{- 0.54041 \dots + 2 π}{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 sin (2 x) + 3 cos (2 x) = 0

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

\frac{5 sin ( 2 x ) + 3 cos ( 2 x )}{cos ( 2 x )} = \frac{0}{cos ( 2 x )}

Simplify

\frac{5 sin ( 2 x )}{cos ( 2 x )} + 3 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

5 tan (2 x) + 3 = 0

5 tan (2 x) + 3 = 0

Move

3

to the right side

5 tan (2 x) + 3 = 0

Subtract

3

from both sides

5 tan (2 x) + 3 - 3 = 0 - 3

Simplify

5 tan (2 x) = - 3

5 tan (2 x) = - 3

Divide both sides by

5

5 tan (2 x) = - 3

Divide both sides by

5

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

tan (x) = - a \Rightarrow x = arctan (- a) + 18 0^{\circ} n

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

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}

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

Simplify

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

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

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{3}{5}) = - arctan (\frac{3}{5})

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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}

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}

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Frequently Asked Questions (FAQ)

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