Solve the equation.
4 sin² x - 3 = 0
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Solve the equation.
4 sin² x - 3 = 0
![4 {sin}^{2} x = 3 \\ \\ {sin}^{2} x = \frac{3}{4} \\ \\ sin \: x = \sqrt{ \frac{3}{4} } \\ \\ sin \: x = ± \frac{ \sqrt{3} }{2} \\ \\ so \\ \\ sin \: x = \frac{ \sqrt{3} }{2} \\ \\ x = {sin}^{ - 1} ( \frac{ \sqrt{3} }{2} ) \\ \\ x = {sin}^{ - 1} ( sin \: ( \frac{\pi}{3} ) \\ \\ = \frac{\pi}{3} \: \: \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ]\\](https://tex.z-dn.net/?f=4+%7Bsin%7D%5E%7B2%7D+x+%3D+3+%5C%5C+%5C%5C+%7Bsin%7D%5E%7B2%7D+x+%3D+%5Cfrac%7B3%7D%7B4%7D+%5C%5C+%5C%5C+sin+%5C%3A+x+%3D+%5Csqrt%7B+%5Cfrac%7B3%7D%7B4%7D+%7D+%5C%5C+%5C%5C+sin+%5C%3A+x+%3D+%C2%B1+%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D+%5C%5C+%5C%5C+so+%5C%5C+%5C%5C+sin+%5C%3A+x+%3D+%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D+%5C%5C+%5C%5C+x+%3D+%7Bsin%7D%5E%7B+-+1%7D+%28+%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D+%29+%5C%5C+%5C%5C+x+%3D+%7Bsin%7D%5E%7B+-+1%7D+%28+sin+%5C%3A+%28+%5Cfrac%7B%5Cpi%7D%7B3%7D+%29+%5C%5C+%5C%5C+%3D+%5Cfrac%7B%5Cpi%7D%7B3%7D+%5C%3A+%5C%3A+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B%5Cfrac%7B+-+%5Cpi%7D%7B2%7D+%2C%5Cfrac%7B%5Cpi%7D%7B2%7D+%5D%5C%5C "4 {sin}^{2} x = 3 \\ \\ {sin}^{2} x = \frac{3}{4} \\ \\ sin \: x = \sqrt{ \frac{3}{4} } \\ \\ sin \: x = ± \frac{ \sqrt{3} }{2} \\ \\ so \\ \\ sin \: x = \frac{ \sqrt{3} }{2} \\ \\ x = {sin}^{ - 1} ( \frac{ \sqrt{3} }{2} ) \\ \\ x = {sin}^{ - 1} ( sin \: ( \frac{\pi}{3} ) \\ \\ = \frac{\pi}{3} \: \: \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ]\\")
![sin \: x = - \frac{ \sqrt{3} }{2} \\ \\ x = {sin}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) \\ \\ x = {sin}^{ - 1} ( sin \: ( - \frac{\pi}{3} ) \\ \\ = - \frac{\pi}{3} \: \: \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2}] \\](https://tex.z-dn.net/?f=sin+%5C%3A+x+%3D+-+%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D+%5C%5C+%5C%5C+x+%3D+%7Bsin%7D%5E%7B+-+1%7D+%28+%5Cfrac%7B+-+%5Csqrt%7B3%7D+%7D%7B2%7D+%29+%5C%5C+%5C%5C+x+%3D+%7Bsin%7D%5E%7B+-+1%7D+%28+sin+%5C%3A+%28+-+%5Cfrac%7B%5Cpi%7D%7B3%7D+%29+%5C%5C+%5C%5C+%3D+-+%5Cfrac%7B%5Cpi%7D%7B3%7D+%5C%3A+%5C%3A+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B%5Cfrac%7B+-+%5Cpi%7D%7B2%7D+%2C%5Cfrac%7B%5Cpi%7D%7B2%7D%5D+%5C%5C+ "sin \: x = - \frac{ \sqrt{3} }{2} \\ \\ x = {sin}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) \\ \\ x = {sin}^{ - 1} ( sin \: ( - \frac{\pi}{3} ) \\ \\ = - \frac{\pi}{3} \: \: \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2}] \\")
Thus there are two values of x ,these are π/3 ,-π/3
4 sin² x - 3 = 0
Thus there are two values of x ,these are π/3 ,-π/3
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