+130536 2sinx+cosx=0 subtract cosx from both sides
2sinx = -cosx square both sides
4sin'^2x = cos^2x and we can write cos^2x as 1 - sin^2x
4sin^2x = 1 - sin^2x add sin^2x to both sides
5sin^2x = 1 divide both side by 5
sin^2x = 1/5 take the positive and negative square roots of both sides
sinx = plus/minus sqrt(1/ 5)
So, taking the sine inverse of both answers, we have
sin-1 ( sqrt (1/5) ) = x = about 26.565 ° and sin-1 (-sqrt (1/5) ) = x = about - 26,565 °
We can reject the first solution because the sin and cos are positive at 26.565 ° ......and this would make the original equation false
The second solution is correct
For more general solutions......here's a graph [ in degrees ] ......https://www.desmos.com/calculator/cttwh9sjr2
One general solution is x = [ - 26.565 ° + 180n ] ° where n is an integer
