Question

i. If 1+ Sinx= 3SinxCosx then the solution
is​

Answer 1

Answer: 1 + sin²x = 3 sinx cosx  

(1 + sin²x)² = 9 sin²x cos²x  

1 + sin⁴x + 2 sin²x = 9 sin²x (1 - sin²x)  

1 + sin⁴x + 2 sin²x = 9sin²x - 9sin⁴x  

10sin⁴x - 7sin²x + 1 = 0  

sin²x = [ -(-7) ± √((-7)² - 4(10)(1))] / 2(10)  

sin²x = [ 7 ± √(49 - 40)] / 20  

sin²x = [ 7 ± 3] / 20  

sin²x = 1/2 or 1/5  

sinx = ±1/√2 or ±1/√5  

tanx  

= sinx/cosx  

= sinx/√(1 - sin²x)  

= (±1/√2)/√(1 - 1/2) OR (±1/√5)/√(1 - 1/5)  

= ±1 OR (±1/√5)/√(4/5)  

= ±1 OR ±1/2