# Solve the trigonometric equations on the interval 0 le theta < 2pi (32) 4(1 + sin theta) = cos^2 theta (38) sec theta = tan theta + cot theta (40)sqrt(3)sin theta + cos theta = 1

Solve the trigonometric equations on the interval $0\le \theta <2\pi$.
(32)$4\left(1+\mathrm{sin}\theta \right)={\mathrm{cos}}^{2}\theta$
(38)$\mathrm{sec}\theta =\mathrm{tan}\theta +\mathrm{cot}\theta$
(40)$\sqrt{3}\mathrm{sin}\theta +\mathrm{cos}\theta =1$
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Osvaldo Crosby
(32) Use ${\mathrm{sin}}^{2}+{\mathrm{cos}}^{2}=1$ to write as
$4\left(1+\mathrm{sin}\theta \right)=1-{\mathrm{sin}}^{2}\theta$
This is a quadratic equation in $\mathrm{sin}\theta :$
${\mathrm{sin}}^{2}\theta +4\mathrm{sin}\theta +3=0$
$\left(\mathrm{sin}\theta +2{\right)}^{2}=1$
$\mathrm{sin}\theta =-1\to \theta =3\pi /2$
(38) Multiply throughout by $\mathrm{cos}\theta :$
$1=\mathrm{sin}\theta +{\mathrm{cos}}^{2}\theta /\mathrm{sin}\theta$
Multiply throughout by $\mathrm{sin}\theta :$
$\mathrm{sin}\theta ={\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta$
Rewrite using ${\mathrm{sin}}^{2}+{\mathrm{cos}}^{2}=1$
$\mathrm{sin}\theta =1\to \theta =\pi /2$
(40) Divide throughout by $\mathrm{cos}\theta :$
$\sqrt{3}\mathrm{tan}\theta +1=\mathrm{sec}\theta$
Square both sides:
$3{\mathrm{tan}}^{2}\theta +2\sqrt{3}\mathrm{tan}\theta +1={\mathrm{sec}}^{2}\theta$
Use ${\mathrm{sec}}^{2}=1+{\mathrm{tan}}^{2}:$
$2{\mathrm{tan}}^{2}\theta +2\sqrt{3}\mathrm{tan}\theta =0$
$\mathrm{tan}\theta =-\sqrt{3}\to \theta =-\pi /3+n\pi$ so here $\theta =2\pi /3$
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Parker Bird
32. $4\left(1+\mathrm{sin}\theta \right)=1-{\mathrm{sin}}^{2}\theta$
so, ${\mathrm{sin}}^{2}\theta +4\mathrm{sin}\theta +3=0$
so, $\left(\mathrm{sin}\theta +1\right)\left(\mathrm{sin}\theta +3\right)=0$
so, $\mathrm{sin}\theta =-1$
so, $\theta =3\pi /2$
38. $1/\mathrm{cos}\theta =\mathrm{sin}\theta /\mathrm{cos}\theta +\mathrm{cos}\theta /\mathrm{sin}\theta$
so, $\mathrm{sin}\theta ={\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$
so, $\theta =\pi /2$
40. $\sqrt{3}\mathrm{sin}\theta +\mathrm{cos}\theta =1$
now, divide both sides by 2
so, $\mathrm{sin}\theta \mathrm{cos}\left(\pi /6\right)+\mathrm{sin}\left(\pi /6\right)\mathrm{cos}\theta =\mathrm{sin}\left(\pi /6\right)$
so, $\mathrm{sin}\left(\theta +\pi /6\right)=\mathrm{sin}\left(\pi /6\right)$
so, $\theta =0,\pi -\pi /6,ie.0,5\pi /6$