Show how '2sin(x)+sec(x+ π/6)=0' can be expressed as √3sin(x)cos(x)+cos^2(x)=0.

To do this question we must first be aware of a few trigonometric identities. 1. sec(x) = 1/cos(x). 2. cos(A + B) = cos(A)cos(B) - sin(A)sin(B). (double angle formula) 3 . sin^2(x) = 1 - cos^2(x) Before diving in, inspect the given expression and expression you are trying to get to. All terms on the RHS have 'x' within the brackets, therefore we must be using double angle formula to separate a sec term on the LHS. Only sin and cos are present on the RHS therefore we must try to change the sec term into another trig function. 1. Using identity 1. above, '2sin(x)+sec(x+ π/6)=0' can be changed to '2sin(x)+1/cos(x + π/6)=0' 2. considering only cos(x + π/6), use identity 2. to express it as functions only containing 1 term. A = x B = π/6 cos(x + π/6) = cos(x)cos(π/6) - sin(x)sin(π/6) here it is useful to notice that cos(π/6) and sin(π/6) are known values. Put each one into your calculator and you will find, cos(π/6) = √3/2 and sin(π/6) = 1/2 therefore 2sin(x)+sec(x+ π/6)=0 → 2sin(x)+1/((√3/2)cos(x) - (1/2)sin(x)) = 0 3. In order to get it into a form closer to the final expression, multiply through by (√3/2)cos(x) - (1/2)sin(x). This gives √3sin(x)cos(x) - sin^2(x) +1=0 4. Using identity 3 above, sin^2(x) → 1 - cos^2(x) simplifying will give √3sin(x)cos(x)+cos^2(x)=0