sin(x+pi/6)+cos(x+pi/4)=0 help!

[green_tea]

Hi!
I need some help with solving this one:
sin(x+pi/6) + cos(x+pi/4)=0

I've come this far:

sin(x)cos(pi/6) + cos(x)sin(pi/6) + cos(x)cos(pi/4) - sin(x)sin(pi/4) = 0

sin(x) * (3^½)/2 + cos(x) * ½ + cos(x) * (1/(2^½)) - sin(x) * (1/(2^½)) = 0

sin(x) * [ (3^½)/2 - 1/(2^½)] + cos(x) * [ (1/ 2^½) + ½] = 0

sin(x) * (6^½ - 2)/(2*2^½) + cos(x) * (2^½ +2)/(2*2^½) = 0

sin(x) * (6^½ - 2) + cos(x) * (2^½ + 2) = 0

And now I'm stuck. I tried to use this formula: A*sin(x) + B*cos(x) = C * sin (x+v), where C = (A^2 + B^2)^½ , cos(v)= A/C, sin(v) = B/C
but then C gets a bit complicated, like (16+2*6^½ + 4*2^½) and I have to solve the equation without calculator, so there has to be a simpler way...
Does anyone know how I should go on with this :?:

 

[galactus]

Here is an identity you can try. May make it easier.

\(\displaystyle sin(x+\frac{\pi}{6})=\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx\)

\(\displaystyle cos(x+\frac{\pi}{4})=\frac{1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx\)

Then we get \(\displaystyle \frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx+\frac{1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx\)

You can whittle this into a tan and solve. Do you need one solution or just one that will suffice?.

 

[Deleted member 4993]

I need some help with solving this one:

sin(x+pi/6) + cos(x+pi/4)=0

another way:

\(\displaystyle \sin(x+\frac{\pi}{6}) \, = \, - \, \cos(x+\frac{\pi}{4})\)

\(\displaystyle \sin(x+\frac{\pi}{6}) \, = \, \cos(\pi + x+\frac{\pi}{4})\)

\(\displaystyle \sin(x+\frac{\pi}{6}) \, = \, \cos( x+\frac{5\pi}{4})\)

\(\displaystyle \sin(x+\frac{\pi}{6}) \, = \, \sin( \frac{\pi}{2}- x-\frac{5\pi}{4})\)

\(\displaystyle \sin(x+\frac{\pi}{6}) \, = \, \sin( -x-\frac{3\pi}{4})\)

\(\displaystyle \sin(x+\frac{\pi}{6}) \, = \, \sin( -x-\frac{3\pi}{4} \pm \, 2n\pi) \, where \, n \, = \, 1, \, 2, \, 3 \, etc\)

\(\displaystyle 2x = \, - \, \frac{11\pi}{12} \, \pm \, 2\cdot n\cdot\pi\)

 

[green_tea]

Aah, so it wasn't so difficult after all... Thanks a lot for helping me out guys!