Question

Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°

Answer 1

`\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|*(-1)\\\\2sin^2y+sin\,y-1=0`

.

`\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_(12)=(-b\pm√(b^2-4ac))/(2a)=(-1\pm√(1-4\cdot2\cdot(-1)))/(2\cdot2)=\\\\=(-1\pm√(1+8))/(4)=(-1\pm√(9))/(4)=(-1\pm3)/(4)\\\\x_1=(-1+3)/(4)=(2)/(4)=\boxed{\bf(1)/(2)}\\\\x_2=(-1-3)/(4)=(-4)/(4)=\boxed{\bf-1}\\\\sin\,y=(1)/(2)\\\\\boxed{\bf y_1=(\pi)/(6)~~or~~(30^o)}\\\\\boxed{\bf y_2=(5\pi)/(6)~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=(3\pi)/(2)~~or~~(270^o)}`