Question

α为锐角,2cos²α+7sinα-5=0,求α度数

快啊，我急需答案，过了今晚就把问题关了

Answer 1

2cos²α+7sinα-5
=2(1-sin²α)+7sinα-5
=2-2sin²α+7sinα-5
=-2sin²α+7sinα-3
=-2(sin²α-7/2sinα+3/2)
=-2(sin²α-7/2sinα+49/16-49/16+3/2)
=-2[(sinα-7/4)^2-49/16+3/2]
=-2[(sinα-7/4)^2-25/16]
=-2(sinα-7/4)^2+25/8
=0
所以(sinα-7/4)^2=25/16
sinα-7/4=5/4或者sinα-7/4=-5/4
sinα=3（舍去）或者sinα=1/2
所以 α等于30度

Answer 2

cos²α=1-sin²α
2cos²α+7sinα-5=2-2sin²α+7sinα-5
=2sin²α-7sinα+3=0
(sinα-3)(2sinα-1)=0
sinα≤1
sinα-3≠0
∴2sinα-1=0
sinα=1/2
a=30

Answer 3

cos²α=1-sin²α带入2（1-sin²α）+7sinα-5=0
解出sinα=3(舍去)or 0.5
锐角，30°