The function f : x → 5 sin2 x+3 cos2 x is deﬁned for the domain 0 ≤ x ≤ π.
(i) Express f(x)in the form a+bsin2 x, stating the values of a and b.
(ii) Hence ﬁnd the values of x for which f(x) = 7 sinx.
(iii) State the range of f

Question

The function f : x → 5 sin2 x+3 cos2 x is deﬁned for the domain 0 ≤ x ≤ π.
(i) Express f(x)in the form a+bsin2 x, stating the values of a and b. 
(ii) Hence ﬁnd the values of x for which f(x) = 7 sinx. 
(iii) State the range of f

dumbcow · Answer

$$\sin^2 x + \cos^2 x = 1$$

abmon98 · Answer

i solved i,ii and iam stuck on iii)

abmon98 · Answer

$$5\sin ^{2}x+3(1-\sin ^{2}x)$$

abmon98 · Answer

Expand to get $$2\sin ^{2}x+3$$

dumbcow · Answer

ok range of sin^2 is [0,1]

so range of f(x) is [3,5]

abmon98 · Answer

why is the range is 3,5

dumbcow · Answer

min value for sin^2 is 0
--> 2(0) +3 = 3

max value for sin^2 is 1
--> 2(1) +3 = 5

abmon98 · Answer

how did you find out that the range of sin^2 is 0 and 1

dumbcow · Answer

range of sin is -1 to 1

the square makes everything positive
any number less than 1 squared is still less than 1
thus
range of sin^2 is 0 to 1