Question

Please help math questions
put number with answers please

Answer 1

See below for answers

Explanation:

Problem 1

`(1)/(sin\theta)=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\(\pi)/(2)+2\pi n=2\theta\\ \\(\pi)/(4)+\pi n=\theta\\ \\\theta=\bigr\{(\pi)/(4),(3\pi)/(4),(5\pi)/(4),(7\pi)/(4)\bigr\}`

Therefore, the second option is correct.

Problem 2

`cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)`

`0=sinx\\\\x=\pi n\\\\x=0,\pi`

`0=2sinx+1\\\\-1=2sinx\\\\-(1)/(2)=sinx\\ \\x=(7\pi)/(6)+2\pi n,(11\pi)/(6)+2\pi n\\\\x=(7\pi)/(6),(11\pi)/(6)`

Therefore, the solution set is
`x=\bigr\{0,\pi,(7\pi)/(6),(11\pi)/(6)\bigr\}`, making the second option correct.

Problem 3

`2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0`

`cosx=0\\\\x=(\pi)/(2)+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ`

`2cosx+1=0\\\\2cosx=-1\\\\cosx=-(1)/(2)\\\\x=(2\pi)/(3)+2\pi n,(4\pi)/(3)+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ`

Therefore, the solution set is
`x=\bigr\{90^\circ,120^\circ,240^\circ,270^\circ\bigr\}`, making the fourth option correct