Question

How do you solve `2-2cos^2x=5sinx+3`?

Answer 1

Solution is `x=210^0 ,x= -30^0`

Explanation:

`2-2cos^2x=5sinx+3 or 2(1-cos^2x)-5sinx-3=0 or 2sin^2x-5sinx-3=0 or 2sin^2x-6sinx+sinx-3=0 or 2sinx(sinx-3)+1(sinx-3)=0 or (2sinx+1)(sinx-3)=0 :. sinx=3 or 2sinx=-1`But `sinx` lies between `[-1,1]` so `sinx != 3 :. 2sinx = -1 or sinx = -1/2`We know `sin(-30)=-1/2 and sin210= -1/2`.So solution is `x=210^0 ,x= -30^0`[Ans]

Answer 2

`(7pi)/6 + 2kpi`
`(11pi)/6 + 2kpi`

Explanation:

`2(1 - cos^2 x) = 5sin x + 3`
`2sin^2 x = 5sin x + 3`--> (because `sin^2 x = 1 - cos^2 x`)
`2sin^2 x - 5sin x - 3 = 0`
Solve this quadratic equation for sin x by the improved quadratic formula in graphic form (Socratic Search)
`D = d^2 = b^2 - 4ac = 25 + 24 = 49` --> `d = +- 7`
There are 2 real roots:
`sin x = - b/(2a) +- d/(2a) = 5/4 +- 7/4 = (5 +- 7)/4`
a. `sin x = 12/4 = 3` (rejected as > 1), and
b. `sin x = -2/4 = -1/2`
Trig table of special arcs and unit circle -->
`sin x = - 1/2` --> `x = - pi/6 and x = - (5pi)/6.`
Their co-terminal arcs are: `(11pi)/6 and (7pi)/6`
General answers
`x = (7pi)/6 + 2kpi`
`x = (11pi)/6 + 2kpi`