Solve the equation 3sin^2 2x= cos^2 2x for 0 <x <180?
回答 (5)
2016-07-31 4:17 pm
0° < x < 180°
Then, 0° < 2x < 360°
3 sin²(2x) = cos²(2x)
sin²(2x)/cos²(2x) = 1/3
tan²(2x) = 1/3
tan(2x) = 1/√3 or tan(2x) = -1/√3
2x = 30°, (180°+30°) or 2x = (180°-30°), (360°-30°)
x = 15°, 75°, 105°, 165°
Then, 0° < 2x < 360°
3 sin²(2x) = cos²(2x)
sin²(2x)/cos²(2x) = 1/3
tan²(2x) = 1/3
tan(2x) = 1/√3 or tan(2x) = -1/√3
2x = 30°, (180°+30°) or 2x = (180°-30°), (360°-30°)
x = 15°, 75°, 105°, 165°
2016-07-31 4:11 pm
3sin^2 2x = cos^2 2x
3sin^2 2x / cos^2 2x = 1
3tan^2 2x = 1
tan^2 2x = 1/3
tan 2x = ± (√3)/3
2x = kpi ± pi/6
x = kpi/2 ± pi/12 , k any integer
for k = 0 and k = 1 and k = 2, answers will be on (0°, 180°) :
x = pi/12, 5pi/12, 7pi/12, 11pi/12
Answers on angles mode :
x = 15°, 75°, 105° and 165°
3sin^2 2x / cos^2 2x = 1
3tan^2 2x = 1
tan^2 2x = 1/3
tan 2x = ± (√3)/3
2x = kpi ± pi/6
x = kpi/2 ± pi/12 , k any integer
for k = 0 and k = 1 and k = 2, answers will be on (0°, 180°) :
x = pi/12, 5pi/12, 7pi/12, 11pi/12
Answers on angles mode :
x = 15°, 75°, 105° and 165°
2016-07-31 4:23 pm
Alternative method
3sin²(2x) = 1 - sin²(2x)
4sin²(2x) = 1
sin²(2x) = 1/4
sinx = ±1/2
2x = 30°, 150°, 210°, 330°
x = 15°, 75°, 105°, 165°
3sin²(2x) = 1 - sin²(2x)
4sin²(2x) = 1
sin²(2x) = 1/4
sinx = ±1/2
2x = 30°, 150°, 210°, 330°
x = 15°, 75°, 105°, 165°
2016-07-31 4:37 pm
4 sin² (2x) = 1
sin² (2x) = 1 / 4
sin (2x) = ± 1 / 2
2x = 30⁰ , 150⁰ , 390⁰ , 510⁰
x = 15⁰ , 75⁰ , 195⁰ , 255⁰
x = 15⁰ , 75⁰_____for given range of x
sin² (2x) = 1 / 4
sin (2x) = ± 1 / 2
2x = 30⁰ , 150⁰ , 390⁰ , 510⁰
x = 15⁰ , 75⁰ , 195⁰ , 255⁰
x = 15⁰ , 75⁰_____for given range of x
2016-07-31 4:12 pm
sin(2x)^2 / cos(2x)^2 = 1/3
tan(2x)^2 = 3/9
tan(2x) = +/- sqrt(3) / 3
2x = 30 + 180 * k , -30 + 180 * k
x = 15 + 90k , -15 + 90k
x = 15 , 105 , 75 , 165
tan(2x)^2 = 3/9
tan(2x) = +/- sqrt(3) / 3
2x = 30 + 180 * k , -30 + 180 * k
x = 15 + 90k , -15 + 90k
x = 15 , 105 , 75 , 165
收錄日期: 2021-04-20 16:25:59
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