Question
$$5 \sin ^{2} x-3 \sin x \cos x-36 \cos ^{2} x>0$$
Step 1
Step 1: First, we divide the given equation by $\cos^2x$ to get it in terms of $\tan x$: $$ \frac{5 \sin ^{2} x}{\cos^2x}-\frac{3 \sin x \cos x}{\cos^2x}-36 >0 $$ which simplifies to $$ 5 \tan ^{2} x-3 \tan x-36 >0 $$ Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 55 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$2 \cos ^{2} x-3 \sin x \cos x+5 \sin ^{2} x=3$
$$ 4 \sin ^{4} 3 x-3 \cos x+5=0 $$
$3 \sin ^{2} x+3 \sin x \cos x-6 \cos ^{2} x=0$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD