+128460 50. 2sinθcosθ + √2 cos θ = 0 factor out cosθ
cosθ (2sin θ + √2 ) = 0
So...either cosθ = 0 and this happpens at pi/2 and 3pi/2 ......or....
2sin θ + √2 = 0 subtract √2 from both sides
2 sinθ = -√2 divide both sides by 2
sinθ = -√2 / 2 and this happens a 5pi/4 and at 7pi/4
So...the answers are pi/2 , 3pi/2./ 5pi/4 and 7pi/4 ⇒ the third answer
51. √(7x) [ √x - 7√7 ] =
√(7) x- 49√x
