Question 747138
{{{(sin(x))^2+tan(x)-1=0}}} <--> {{{1-(cos(x))^2+tan(x)-1=0}}} <--> {{{-(cos(x))^2+tan(x)=0}}} <--> {{{(cos(x))^2=tan(x)}}} <--> {{{(cos(x))^2=sin(x)/cos(x)}}} <--> {{{(cos(x))^3=sin(x)}}}
{{{(cos(x))^3=sin(x)}}} --> {{{(cos(x))^6=(sin(x))^2}}} --> {{{(cos(x))^6=1-(cos(x))^2}}}
Calling {{{(cos(x))^2=y}}} we can re-write that equation as
{{{y^3=1-y}}} <--> {{{y^3+y-1=0}}}
The only real solution is approximately {{{y=0.6823278038}}}
Going back to {{{x}}} we have {{{(cos(x))^2=0.6823278038}}}
Between {{{0^o}}} and {{{360^o}}} (between {{{0}}} and {{{2pi}}} radians) there are 4 angles with {{{(cos(x))^2=0.6823278038}}}.
However, solutions to {{{(cos(x))^2=tan(x)}}} require positive tangent, so the only solutions will be in the first and third quadrants.
{{{x=34.3^o}}} (or {{{x=0.598767}}} radians) is the approximate solution in the first quadrant.
{{{x=34.3^o+180^o=214.3^o}}} (or {{{x=0.598767+pi=3.740539}}} radians) is the approximate solution in the third quadrant.