Question

Solve the equation for the interval [0, 2π).
cos^2x + 2 cos x + 1 = 0   

2 sin^2x = sin x

cos x = sin x

sec^2x - 2 = tan^2x

Answer 1

Cosx = COST x = X cos tacosxt1=0 cose + cosa + cosx + 1 =0 COLIC Code + 1) + 1 (cosse +1)=0 Cosx + 1)² = 0 COOK+1=0 Cosa = -1 & X=25-T , = asine sina sina 2 sinx=1 Sina = 1/2 Sesx = sin (7/6) x = 1/6 gives the values is quadrant II X-T76 67- 2 Sina sin T-X = 6 - SN/6

3 lan Cosa = Susa COS Cosa tanx = tana tas (774) x = 7/4 is tve is quadrant D, atx AT 7/4 4THT = ST 4 secx - 2 = tantx It tant-2= torta - 1=0 2 4 seéx=+taris Therefore, this equation is not posseble.