Trigonomania!
If the equation above possesses a solution for real , then can take values only between .
Find .
The answer is 8.
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1 solution
a s i n x + c o s ( 2 x ) = 2 a − 7 a s i n x + ( 1 − 2 s i n 2 x ) = 2 a − 7 2 s i n 2 x − a s i n x + 2 a − 8 = 0 S o l v i n g f o r t h e q u a d r a t i c , w e f i n d s i n x = 4 a ± a 2 − 8 ( 2 a − 8 ) = 4 a ± ( a − 8 ) s i n x = 2 a − 4 . { I g n o r i n g t h e o t h e r s o l u t i o n b e c a u s e s i n x = 2 i s n o t p o s s i b l e . } E q u a t i o n h a s s o l u t i o n o n l y i f − 1 ≤ 2 a − 4 ≤ 1 { S i n c e s i n x ϵ [ − 1 , 1 ] } T h e r e f o r e , 2 ≤ a ≤ 6 x = 2 , y = 6 . x + y = 8 .