1. ## solving this equation

sin x - square root of 3 x cos x=2

no idea how to do it o.o

2. Originally Posted by alessandromangione
sin x - square root of 3 x cos x=2

no idea how to do it o.o
This is ambigious. Please pick the correct one :

$1) \ \ \sin{(x)} - \sqrt{3x \cos{(x)}} = 2$

$2) \ \ \sin{(x)} - \sqrt{3x} \cos{(x)} = 2$

$3) \ \ \sin{(x)} - \sqrt{3} x \cos{(x)} = 2$

$\ \ \sin{(x)} - \sqrt{3} \times \cos{(x)} = 2$

4. Lol, didn't see the middle x was slightly shorter than the other ones
Actually it isn't. So it still was ambigious

To answer the question : consider putting all trigonometric functions on one side, put the square root and the constant on the other, and see what you can work out.

$\sin{(x)} - \sqrt{3} \times \cos{(x)} = 2$

Use the trigonometric identity :

$\sin{(x)} - \sqrt{3} \times \sqrt{1 - \sin^2{(x)}} = 2$

Now arrange the equation :

$\sin{(x)} - 2 = \sqrt{3} \times (1 - \sin^2{(x)})$

Square both sides :

$(\sin{(x)} - 2)^2 = 3 \times (1 - \sin^2{(x)})$

Rearrange this :

$(\sin{(x)} - 2)^2 = 3 - 3 \sin^2{(x)}$

Expand on the left hand side :

$\sin^2{(x)} - 4 \sin{(x)} + 4 = 3 - 3 \sin^2{(x)}$

Rearrange :

$\sin^2{(x)} - 4 \sin{(x)} + 3 \sin^2{(x)} = 3 - 4$

Rearrange :

$4 \sin^2{(x)} - 4 \sin{(x)} = -1$

Keep going by factorizing :

$4 ( \sin^2{(x)} - \sin{(x)}) = -1$

Finish this up :

$\sin^2{(x)} - \sin{(x)} = \frac{-1}{4}$

And ... MAGIC ! Quadratic equation :

$\sin^2{(x)} - \sin{(x)} + \frac{1}{4} = 0$

Can you finish now ? It should be fairly easy.

5. thank uou so much for the help!