Question

$\sqrt{3} cosx+sinx=\sqrt{2}$

Answer 1

√3 cosx +sinx = √2

√3 cosx = √2 -sinx

Now square both sides,

3 cos²x = 2 + sin²x -2√2 sinx

3(1-sin²x) = 2 + sin²x -2√2 sinx

3-3sin²x = 2 + sin²x -2√2 sinx

0 = -1 + 4sin²x -2√2 sinx

4sin²x - 2√2 sinx - 1 = 0

let y = sinx, then,

4y² -2√2y -1 =0

y = {2√2±√(8+16)}/8

y = {2√2 ±2√6}/8

y = {√2 ±√6}/4

sinx = {√2 ±√6}/4

x = sin inverse {√2 ±√6}/4