solve√3cosx+sinx=√2
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Sia ? 6 years, 1 month ago
√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
2Thank You