Verified answer

√3cosx + sinx =√3

sin x = √3(1 - cosx)

squaring on both sides

sin^2(x) = 3[ 1 + cos^2(x) - 2cos x ]

1 - cos^2(x) = 3 + 3cos^2(x) - 6cos x

=> 4cos^2(x) - 6cos x + 2 = 0

=> 2cos^2(x) - 3cos x + 1 = 0

=> (2cos(x) - 1)(cos x - 1) = 0

2cos x = 1 and cos x = 1

cos x = 1/2

x = - Π/3 and Π/3

but -Π/3 is extraneous

cos x = 1

x = 0

but x = -Π/3 is extraneous

so x = 0 and Π/3

You MUST get rid of one of the trigonometric functions. i.e. You cannot have both sine and cosine in the equation. First, subtract sqrt.(3) Cos X to isolate Sin X. Then, square both sides to get:

Sin^2(X)= ((Sqr.3) - (Sqrt.(3) Cos(X)))^2. Sin^2(X)= 1-Cos^2(X).

1-Cos^(X)= 3 Cos^2(X) - 6Cos(X) + 3. Put this in standard form.

2Cos^2(X) - 6Cos(X) + 4=0 Factor out the GCF.

2 (Cos^2(X) - 3Cos(X) + 2)=0, 2 divides out.

(Cos X - 2)(Cos X -1) =0

Cos X =2 and Cos X=1

Undefined X=0 pi.

X= 0 pi

x = 0 is one answer