Question

The number of values of triplet (a, b, c) for which acos2x+bsin^2x+c=0 is satisfied by all real x is

Answer 1

we have to find the number of values of triplet (a , b , c) for which acos²x + bsin²x + c = 0 is satisfied by all real value of x.

solution : acos²x + bsin²x + c = 0

⇒acos²x + bsin²x = -c

if we compare it trigonometric identity, sin²x + cos²x = 1

we get, a = b = -c

if we put, a = 1 then, b = 1, c = -1

then, sin²x + cos²x = 1 , satisfied

if we put, a = 2 then, b = 2 , c = -2

2cos²x + 2sin²x - 2 = 0

⇒cos²x + sin²x = 1 , satisfied

if we put, a = m, then, b = m , c = -m

mcos² + msin²x + (-m) = 0

⇒cos²x + sin²x = 1 , satisfied

now you see, it is clearly shown that we can assume any real value of (a, b, c) [ just have to follow condition a = b = - c] for which a cos²x + bsin²x + c = 0 is satisfied by all real x.

Therefore the infinite number of triplet (a, b , c) for which acos²x+bsin²x+c=0 is satisfied by all real x.

also read similar questions : If three unequal positive real numbers satisfying b^2=ac and b – c, c – a, a – b are in H.P. then the value of a + b + c...

https://brainly.in/question/17981338

in triangle abc prove that asin(B-C)+bsin(C-A)+csin(A-B)=0

https://brainly.in/question/3518633