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

2 Answers
Konstantinos Michailidis
Sep 22, 2015

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.

Nghi N.
Sep 22, 2015

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#

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