Question 1006455
.
-3cos(pi/2 - X) = tanX
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<pre>
cos({{{pi/2 - X}}}) = sin(X), according to reduction formula.

Therefore, your equation takes the form

-3*sin(X) = tan(X).   

Rewrite it in the form 

-3*sin(X) = {{{(sin(X))/(cos(X))}}}, or, even better :)

-3*sin(X) = {{{(sin(X))/(sqrt(1-sin^2(X)))}}}

Now introduce the new variable u = sin(X) and square both sides of the last equation. You will get an equation for u:

{{{9*u^2}}} = {{{u^2/(1-u^2)}}}.

Simplify it and solve step by step:

{{{9*u^2*(1-u^2)}}} = {{{u^2}}}  ----->  {{{9*u^4 - 9*u^2}}} = {{{-u^2}}}  ----->  {{{9*u^4 - 8*u^2}}} = {{{0}}}  ----->  {{{u^2*(9*u^2 - 8)}}} = {{{0}}}.

The last equation comes apart in two equations. First one is 

{{{u^2}}} = {{{0}}}  ----->  sin(X) = 0  ----->  X = 0, +- {{{pi}}}, +/- {{{2pi}}}, . . . , +/- {{{k*pi}}}, . . . , k= 0, 1, 2, 3, . . . 

The second one is 

{{{9*u^2-8}}} = {{{0}}}  ----->  {{{u^2}}} = {{{8/9}}}  ----->  {{{u}}} = +/- {{{2*sqrt(2)/3}}}  ----->  sin(X) = +/- {{{2*sqrt(2)/3}}}.  

It generates two families of potential roots: 

(a) X = +/- arcsin({{{(2*sqrt(2))/3}}}) + {{{2k*pi}}}, k = 0, +/-1, +/-2, +/-3, . . . and 

(b) X = +/- [{{{pi - arcsin((2*sqrt(2))/3)}}}] + {{{2k*pi}}}, k = 0, +/-1, +/-2, +/-3, . . .

<TABLE> 
  <TR>
  <TD> 

&nbsp;&nbsp;&nbsp;&nbsp;{{{graph( 330, 330, -8.5, 8.5, -5.5, 5.5,
          -3*sin(x), tan(x)
)}}}


<U>Figure</U>. Plots -3*sin(x) (in red) and tan(x) (in green)

  </TD>
  </TR>
</TABLE>The roots (a) X = +/- arcsin({{{(2*sqrt(2))/3}}}) + {{{2k*pi}}} are excessive. They are not the solutions.

The roots (b) X = +/- [{{{pi - arcsin((2*sqrt(2))/3)}}}] + {{{2k*pi}}} are the solutions.
</pre>