Question

if cos^6a + sin^6a + 2Ksin2a=1, then K=

Answer 1

Step-by-step explanation:

(Cos²a)³+ (Sin²a)³ + 2kSin2a = 1

(Cos²a+Sin²a){Cos⁴a - Cos²a*Sin²a + Sin⁴a}+ 2kSin2a = 1

(Cos²a + Sin²a)² - 3Sin²a*Cos²a + 2kSin2a = 1

1 - 3Sin²a*Cos²a + 2kSin2a = 1

2K Sin2a = 3Sin²a*Cos²a

4k*Sina*Cosa = 3Sin²a*Cos²a

4k = 3Sina*Cosa

k = 3/4 * Sina*Cosa

k = (3/8)Sin2a