Question

lf `sqrt3costheta + sintheta = sqrt2`, then the general value of θ is ______ 

Options

• `npi + (-1)^n pi/4`

• `(-1)^n pi/4 - pi/3`

• `npi + pi/4 - pi/3`

• `npi + (-1)^n pi/4 - pi/3`

Answer 1

lf `sqrt3costheta + sintheta = sqrt2`, then the general value of θ is `underline(npi + (-1)^n pi/4 - pi/3)`.

Explanation:

`sqrt3costheta + sintheta = sqrt2`

D.ividing both sides by `sqrt((sqrt3)^2 + 1^2) = 2`, we get

`sqrt3/2 costheta + 1/2 sintheta = sqrt2/2`

⇒ sinθ cos`pi/3` + cosθ sin`pi/3 = 1/sqrt2`

⇒ `sin(theta + pi/3) = sin(pi/4)`

⇒ `theta + pi/3 = npi + (-1)^n pi/4`

⇒ `theta = npi + (-1)^n pi/4 - pi/3`