Question

How do you find the general solutions for `sin(a)-sqrt3*cos(a)=2`?

Answer 1

Refer to explanation

Explanation:

We have that

`sina-sqrt3*cosa=2=>1/2*sina-sqrt3/2*cosa=1=> cos(pi/6)*cosa-sin(pi/6)*sin(a)=-1=> cos(a+pi/6)=-1=>cos(a+pi/6)=cospi`

Hence we `a+pi/6=2kpi+pi=>a=2kpi+(5*pi)/6` and `a+pi/6=2kpi-pi=>a=2kpi-(7pi)/6`

where `k` is an integer.

Answer 2

Solve `sin x - sqrt3cos x = 2`

Ans:`((5pi)/6)`

Explanation:

`Call tan a = sin a/cos a = sqrt3 = tan (pi/3).`
Apply the trig identity: sin (a - b) = sin a.cos b - sin b.cos a.
The equation becomes:
`sin x - (sin a/cos a)cos x = 2`
`sin x cos a - sin a.cos x = 2cos a`
`sin (x - a) = sin (x - pi/3) = 2cos ((pi)/3)= 1`
`sin (x - pi/3) = 1 = sin ((pi)/2)`
`x - pi/3 = pi/2` --> `x = ((pi)/2) + ((pi)/3) = ((5pi)/6)`
General solution `x = ((5pi)/6)` + `2kpi`