Answer:
[tex]\cos(x-0.523) = \frac{1}{2}[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]h(x) = \sqrt{3} \sin(x) + \cos(x)[/tex]
We have to write this in
[tex]h(x) = R\cos (x-c)[/tex]
We can do this in the following manner:
[tex]a = R\cos c\\b = R\sin c\\\text{Putting vallues we get}\\h(x) = a\sin x + b\cos x\\= R\cos c \sin x + R \sin c \cos x\\=R\cos(x-c)\\R = \sqrt{a^2 + b^2}\\\\\tan c = \displaystyle\frac{b}{a}[/tex]
Thus, we can write:
[tex]R = \sqrt{\sqrt{3}^2 + 1^2} = \sqrt{3+1} = 2\\h(x) = R\cos(x-c)\\h(x) = 2\cos(x-c)\\\\\tan c = \frac{1}{\sqrt{3}}\\\\c = \tan^{-1}\frac{1}{\sqrt{3}} = 0.523\text{ radians}\\\\\cos(x-0.523) = \frac{1}{2}[/tex]