Trigonometric Ratios and Identities

Given, 3‌sin‌x+4‌cos‌x=k+1. . . (i)
Multiply and divide LHS of Eq. (i) by √32+42=5
i.e. 5(‌

3

5

‌sin‌x+‌

4

5

‌cos‌x)=k+1
⇒5(cos‌α‌sin‌x+sin‌α‌cos‌x)=k+1
[Let cos‌α=3∕5 then sin‌α=√1−(3∕5)2=‌

4

5

]
⇒5‌sin(x+α)=k+1‌‌[Use‌sin(a+b)=sin‌a‌cos‌b+cos‌a‌sin‌b]
⇒sin(x+α)=‌

k+1

5

Let x+α=θ
Then, sin‌θ=‌

k+1

5

∵‌‌−1≤sin‌θ≤1
⇒‌‌−1≤‌

k+1

5

≤1
⇒‌‌−5≤k+1≤5
⇒‌‌−6≤k≤4
∴ Possible integral values of k are −6,−5,−4,−3,−2,−1,0,1,2, 3 and 4. i.e.
Total 11 integral values of k are possible for which Eq. (i) has solution.