solve√3cosx+sinx=√2

1.	solve√3cosx+sinx=√2
Answer» √3 cosx +sinx = √2√3 cosx = √2 -sinxNow square both sides,3 cos²x = 2 + sin²x -2√2 sinx3(1-sin²x) = 2 + sin²x -2√2 sinx3-3sin²x = 2 + sin²x -2√2 sinx0 = -1 + 4sin²x -2√2 sinx4sin²x - 2√2 sinx - 1 = 0let y = sinx, then,4y² -2√2y -1 =0y = {2√2±√(8+16)}/8y = {2√2 ±2√6}/8y = {√2 ±√6}/4sinx = {√2 ±√6}/4x = sin inverse {√2 ±√6}/4

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