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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
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\[\sin^2 x + \cos^2 x = 1\]
i solved i,ii and iam stuck on iii)
\[5\sin ^{2}x+3(1-\sin ^{2}x)\]
Expand to get \[2\sin ^{2}x+3\]
ok range of sin^2 is [0,1] so range of f(x) is [3,5]
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why is the range is 3,5
min value for sin^2 is 0 --> 2(0) +3 = 3 max value for sin^2 is 1 --> 2(1) +3 = 5
how did you find out that the range of sin^2 is 0 and 1
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
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