Question 1058139
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Find the general solution √3cosx+sinx=1
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<pre>
{{{sqrt(3)*cos(x) + sin(x)}}} = {{{1}}}.

Multiply both sides by {{{1/2}}}. You will get

{{{(sqrt(3)/2)*cos(x) + (1/2)*sin(x)}}} = {{{1/2}}}.           (1)

Notice that {{{sqrt(3)/2}}} = {{{sin(pi/3)}}}, {{{1/2}}} = {{{cos(pi/3)}}}. 

Substitute it into the left side of (1). You will get

{{{sin(pi/3)*cos(x) + cos(pi/3)*sin(x)}}} = {{{1/2}}}.    (2)

Apply the formula sin(a)*cos(b) + cos(a)*sin(b) = sin(a+b) to the left side of (2). You ill get

{{{sin(pi/3 + x)}}} = {{{1/2}}}.                       (3)

It implies 

{{{pi/3 + x}}} = {{{pi/6 + 2k*pi}}},  k = 0, +/-1, +/-2, . . . or

{{{pi/3 + x}}} = {{{5pi/6 + 2k*pi}}},  k = 0, +/-1, +/-2, . . .


Thus there are two sets of solutions:

1.  x = {{{pi/6 - pi/3 + 2k*pi}}} = {{{-pi/6 + 2k*pi}}},  which is equivalent to  x = {{{11pi/6 + 2k*pi}}},


and the other family


2.  x = {{{5pi/6 - pi/3 + 2k*pi}}} = {{{pi/2 + 2k*pi}}}


<U>Answer</U>.  There are two sets of solutions:  1)  x = {{{11pi/6 + 2k*pi}}}  and  2)  x = {{{pi/2 + 2k*pi}}},  k = 0, +/-1, +/-2, . . .
</pre>

The plot below confirms these solutions.


{{{graph( 330, 330, -0.5, 6.5, -2.5, 2.5,
          sqrt(3)*cos(x) + sin(x), 1
)}}}


Plots y = {{{sqrt(3)*cos(x) + sin(x)}}} and y = 1