Question

Trigonometry problem​

Answer 1

The values are x = 0° or x = 60°

Explanation:

Given:

`\cos x + (1)/(√(3) )* \sin x = 1`

Let it be in the form of

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

a = 1

`b=(1)/(√(3) )`

We have

`\sqrt{(a^(2)+b^(2) )}=\sqrt{(1^(2)+((1)/(√(3) )) ^(2) )}\\=\sqrt{(4)/(3)}\\=(2)/(√(3) )`

Now, Dividing both the side by
`(2)/(√(3) )` we get

`(√(3) )/(2)\cos x + (1)/(2)\sin x =(√(3) )/(2)\\`

We Know

`\sin 60 =(√(3) )/(2)\\and\\\cos 60 = (1)/(2)`

Now by replacing with above values we get

`\sin 60* \cos x + \cos 60* \sin x = (√(3) )/(2)\\`

Also we have formula

`\sin A* \cos B + \cos A* \sin B = \sin (A+B)`

By applying the above Formula we get

`\sin (60+x)=(√(3) )/(2)\\`

Also ,

`\sin (60)=(√(3) )/(2)\\and\\\sin (120)=(√(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°