Math  /  Trigonometry

Question12. Risolvi l'equazione 2cos2(x)3sin(x)1=02 \cos ^{2}(x)-3 \sin (x)-1=0 nell'intervallo [0,2π][0,2 \pi]. (a) x=π4,5π4x=\frac{\pi}{4}, \frac{5 \pi}{4} (b) x=π6,5π6x=\frac{\pi}{6}, \frac{5 \pi}{6} (c) x=π3,2π3,4π3,5π3x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} (d) Nessuna soluzione (e) x=arcsin(13),πarcsin(13)x=\arcsin \left(-\frac{1}{3}\right), \pi-\arcsin \left(-\frac{1}{3}\right)

Studdy Solution
La risposta corretta è (e): x=arcsin(13),πarcsin(13)x = \arcsin(-\frac{1}{3}), \pi - \arcsin(-\frac{1}{3})

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord