Question

√3 sinx +cosx = 1 how do you find x?

Answer 1

use trig identities. look'em up online.

Answer 2

i got up to 3sin^2x +cos^2x+2√3 sinx cosx =1

Answer 3

There is one for this problem. \(a\sin \theta+b\cos \theta=\sqrt{a^2+b^2}\sin(\theta+\theta_1)\) where \[\sin \theta_1=\frac{b}{\sqrt{a^2+b^2}}\].

Answer 4

first you should have square rooted both sides. then subtract sine, then use trig ids.

Answer 5

From the equation \[\sqrt{3} \sin x+\cos x=1\]let \(a=\sqrt{3}\) & \(b=1\). Thus \[\sqrt{3} \sin x+\cos x=\sqrt{(\sqrt{3})^2+1^2}\sin(x+\theta_1)\]where \[\sin \theta_1=\frac{1}{\sqrt{(\sqrt{3})^2+1^2}}=\frac{1}{2}\] So \(\theta_1=\frac{\pi}{6}\). So the given equation reduces to \[2\sin\left(x+\frac{\pi}{6}\right)=1\]

Answer 6

The fun way of doing this is to multiply both sides by 1/2 and noticing that you have the dot product of two vectors, <sinx, cosx> and <sin(pi/3), cos(pi/3)> and the projection of one onto the other is 1/2, so it must be pi/2 away. So, x=pi/3+pi/2.