Answer:
The values are x = 0° or x = 60°
Step-by-step explanation:
Given:
[tex]\cos x + \frac{1}{\sqrt{3} }\times \sin x = 1[/tex]
Let it be in the form of
[tex]a\cos x + b\sin x = 1[/tex] such that
a = 1
[tex]b=\frac{1}{\sqrt{3} }[/tex]
We have
[tex]\sqrt{(a^{2}+b^{2} )}=\sqrt{(1^{2}+(\frac{1}{\sqrt{3} }) ^{2} )}\\=\sqrt{\frac{4}{3}}\\=\frac{2}{\sqrt{3} }[/tex]
Now, Dividing both the side by [tex]\frac{2}{\sqrt{3} }[/tex] we get
[tex]\frac{\sqrt{3} }{2}\cos x + \frac{1}{2}\sin x =\frac{\sqrt{3} }{2}\\[/tex]
We Know
[tex]\sin 60 =\frac{\sqrt{3} }{2}\\and\\\cos 60 = \frac{1}{2}[/tex]
Now by replacing with above values we get
[tex]\sin 60\times \cos x + \cos 60\times \sin x = \frac{\sqrt{3} }{2}\\[/tex]
Also we have formula
[tex]\sin A\times \cos B + \cos A\times \sin B = \sin (A+B)[/tex]
By applying the above Formula we get
[tex]\sin (60+x)=\frac{\sqrt{3} }{2}\\[/tex]
Also ,
[tex]\sin (60)=\frac{\sqrt{3} }{2}\\and\\\sin (120)=\frac{\sqrt{3} }{2}[/tex]
Comparing we get
60 + x = 60 or 60 + x = 120
∴ x = 0° ∴ x = 120 - 60 = 60°
Therefore, the values are x = 0° or x = 60° i.e between 0° to 360°
