giải pt a.4sin ²2x+8cos ²x-5=0 b.-4cos ²x+8cos2x+2=0 ai giúp e vs ạ cảm ơn 26/07/2021 Bởi Aaliyah 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` Bình luận
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`
⇔−2−2cos2x+8cos2x+ 2=0⇔-2-2cos2x+8cos2x+ 2=0
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