on the interval of [0,π] solve tan^2 x+3secx= −3


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Question Number 107978 by anonymous last updated on 13/Aug/20

$o n t h e i n t e r v a l o f [0, π] s o l v e$ $\tan^{2} x + 3 s e c x = - 3$
	Answered by bemath last updated on 14/Aug/20
	$((•Bemath•)/(Cooll)) ⇒sec^2 x+3sec x+3−1=0 ⇒sec^2 x+3sec x+2=0 { ((sec x=−1⇒cos x=−1)),((sec x=−2⇒cos x=−(1/2))) :}$
	$\frac{∙ B e m a t h ∙}{C o o l l}$ $\Rightarrow \sec^{2} x + 3 \sec x + 3 - 1 = 0$ $\Rightarrow \sec^{2} x + 3 \sec x + 2 = 0$ ${\begin{cases} \sec x = - 1 \Rightarrow \cos x = - 1 \\ \sec x = - 2 \Rightarrow \cos x = - \frac{1}{2} \end{cases}$
	Commented by anonymous last updated on 13/Aug/20

	$T h a n k a l o t s i f$
	Commented by aleks041103 last updated on 13/Aug/20

	$S h o u l d b e c o s x = - 1$
	Commented by bemath last updated on 14/Aug/20

	$y e s . . . . c o o l l$
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