giải pt a.4sin ²2x+8cos ²x-5=0 b.-4cos ²x+8cos2x+2=0 ai giúp e vs ạ cảm ơn

a) `4sin^2 2x+8cos^2x-5=0`

`<=>  4(1-cos^22x)+8. {1+cos2x}/{2}-5=0`

`<=> -4cos^2 2x+ 4 +4 (1+cos2x)-5=0`

`<=> 4cos^2 2x- 4cos2x -3=0 `

`<=> cos2x = -1/2 (TM)` hoặc `cos2x = 3/2 (L)`

`<=> cos2x = cos (2π)/3`

<`=> 2x = \pm (2π)/3 + k2π`

`<=> x = \pm π/3 +kπ`

b) `-4cos^2x + 8cos2x + 2=0`

`<=> -2(1+cos2x) + 8cos2x + 2 =0`

`<=> -2 – 2cos2x + 8cos2x +  2 =0`

`<=> cos2x =0`

`<=> 2x = π/2 + kπ`

`<=> x = π/4 + (kπ)/2`