Question 1049302
.
Tangent (theta) + square root of 3 = secant (theta)


I tried squaring both sides but in the end I never result with the correct answer. 
I'm not sure how to start this one or what I am doing wrong.
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<pre>
So, you are trying to solve an equation

tan(x) + {{{sqrt(3)}}} = sec(x).

It is the same as

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

Multiply both sides by cos(x). You will get

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


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

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


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

Therefore, you can write the last equation as 

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


Apply the addition formula for sine.  ( It is cos(a)*sin(b) + sin(a)*cos(b) = sin(a+b). 
                                        See the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-formulas.lesson>Addition and subtraction formulas</A> in this site ). You will get

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

It implies  {{{x + pi/3}}} = {{{pi/6}}}  or  {{{x + pi/3}}} = {{{5pi/6}}}.

Hence,  x = {{{pi/6-pi/3}}} = {{{-pi/6}}}  or  x = {{{5pi/6-pi/3}}} = {{{3pi/6}}} = {{{pi/2}}}.

The last root doesn't fit due to "sec" in the original equation.


<U>Answer</U>.  x = {{{-pi/6}}},  or  {{{-pi/6 + 2k*pi}}} for any integer "k".
</pre>

See also the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Solving-simple-problems-on-trigonometric-equations.lesson>Solving simple problems on trigonometric equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Solving-more-complicated-problems-on-trigonometric-equations.lesson>Solving more complicated problems on trigonometric equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Solved-problems-on-trigonometric-equations.lesson>Solving advanced problems on trigonometric equations</A>

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