the equation: a sinx+cos2x=2a-7possess a sol.,if?

Question

choose
1)a>6
2)2<=a<=6
plese xplain ur ans

Answer 1

asinx+(1-2sin^2x)=2a-7
asinx+1-2sin^2x=2a-7
-2sin^2x+asinx+1-2a+7=0
-2sin^2x+asinx-2a+8=0
-2sin^2x+asinx-(2a-8)=0
use the quadratic formula
here a=-2;b=a and c=-(2a-8)
condition for real solution is
b^2>4ac
=>a^2>(-8){-(2a-8)}
=>a^2>(-8)(-2a+8)
=>a^2>16a-64
=>a^2-16a+64>0
=>(a-8)^2=0
(b^2-4ac)^1/2=(a-8)
x1=(-a+(a-8))/-4
x2=(-a-(a-8))/-4
x1not admissible
x2 has to be bet -1 and +1
i.e. -2a+8/-4bet -1 and +1
-2(a-4)/-4 must be bet -1 and +1
a-4 must be bet -2 and +2
a must be between 2 and 6

Answer 2

cos 2x = 1 - (sinx)^2
so asinx + 1-(sinx)^2=2a - 7
we replace sinx with t and get
at+1-t^2-2a+7=0
t^2-at+2a-8=0
you solve the eq with delta = b^2-4*a*c, !!! where a,b,c are the coefficients in the eq and not your parameter !!!
and then t=(a(+/-)(a^2-8a+32)^1/2)/2, (here is the parameter)
but t is sinx which can only be between -1 and 1
so (a)(+/-)(a^2-8a+32)^1/2 between -2 and 2
hope you understand eq 2 grade
(+/-) means plus/minus that's both at the same time and there is a square root there hope you see it!)
so you go by turns
first up
(a)(+/-)(a^2-8a+32)^1/2 <2
a-2 < that square root
you square everything
a^2 - 4a+4 which means a<7 or a<= 6
second you do the same for (a)(+/-)(a^2-8a+32) > -2
and the answer is number 2

Answer 3

The way I did this was to plug in values for a and try to solve for x. When I put in 7 for a, the only solutions for x were imaginary. However, when I put in 3 for a, there was at least one value for x that made the equation true. The answer itself is not as important as the fact that there is one at all. So, your two ansers are:
1)No.
2)Yes!

Answer 4

ask ur teachers

Answer 5

I'm sorry. Could you please repeat the question??