Question

Trigonometry problem​

Answer 1

Answer:

The values are x = 0° or x = 60°

Step-by-step explanation:

Given:

$\cos x + \frac{1}{\sqrt{3} }\times \sin x = 1$

Let it be in the form of

$a\cos x + b\sin x = 1$ such that

a = 1

$b=\frac{1}{\sqrt{3} }$

We have

$\sqrt{(a^{2}+b^{2} )}=\sqrt{(1^{2}+(\frac{1}{\sqrt{3} }) ^{2} )}\\=\sqrt{\frac{4}{3}}\\=\frac{2}{\sqrt{3} }$

Now, Dividing both the side by $\frac{2}{\sqrt{3} }$ we get

$\frac{\sqrt{3} }{2}\cos x + \frac{1}{2}\sin x =\frac{\sqrt{3} }{2}$

We Know

$\sin 60 =\frac{\sqrt{3} }{2}\\and\\\cos 60 = \frac{1}{2}$

Now by replacing with above values we get

$\sin 60\times \cos x + \cos 60\times \sin x = \frac{\sqrt{3} }{2}$

Also we have formula

$\sin A\times \cos B + \cos A\times \sin B = \sin (A+B)$

By applying the above Formula we get

$\sin (60+x)=\frac{\sqrt{3} }{2}$

Also ,

$\sin (60)=\frac{\sqrt{3} }{2}\\and\\\sin (120)=\frac{\sqrt{3} }{2}$

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°