What is the general solution for -13a*sin(2x)-13b*cos(2x)=0 ?

The general solution for -13a*sin(2x)-13b*cos(2x)=0 is x=(arctan(-b/a))/2+(pin)/2

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-13asin(2x)-13bcos(2x)=0

- 13 a \cdot sin (2 x) - 13 b \cdot cos (2 x) = 0

x = \frac{arctan ( - \frac{b}{a} )}{2} + \frac{πn}{2}

Solution steps

- 13 a sin (2 x) - 13 b cos (2 x) = 0

Rewrite using trig identities

- 13 a tan (2 x) - 13 b = 0

Move

13 b

to the right side

- 13 a tan (2 x) = 13 b

Divide both sides by

- 13 a; a \neq = 0

tan (2 x) = - \frac{b}{a}; a \neq = 0

Apply trig inverse properties

2 x = arctan (- \frac{b}{a}) + πn

Solve

2 x = arctan (- \frac{b}{a}) + πn : x = \frac{arctan ( - \frac{b}{a} )}{2} + \frac{πn}{2}

x = \frac{arctan ( - \frac{b}{a} )}{2} + \frac{πn}{2}

3 + sin (x) = 2

1 - sin (A) cos (A) = (sin (A) - cos (A))^{2}

sin (43. 5^{\circ}) = 1.5 \cdot sin (x)

cos (x) = \frac{- 1}{10}

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

What is the general solution for -13a*sin(2x)-13b*cos(2x)=0 ?
The general solution for -13a*sin(2x)-13b*cos(2x)=0 is x=(arctan(-b/a))/2+(pin)/2