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Ответ:

[tex] { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1 \\ {( \frac{3}{4} )}^{2} + { \cos( \alpha ) }^{2} = 1 \\ \frac{9}{16} + { \cos( \alpha ) }^{2} = 1 \\ { \cos( \alpha ) }^{2} = 1 - \frac{9}{16} \\ { \cos( \alpha ) }^{2} = \frac{16 - 9}{16} \\ { \cos( \alpha ) }^{2} = \frac{7}{16} \\ \cos( \alpha ) = + - \sqrt{ \frac{7}{16} } \\ \cos( \alpha ) = + - \frac{ \sqrt{7} }{4} [/tex]

Объяснение:

[tex]sin\alpha =\frac{3}{4} \ \ \ \ \ cos\alpha =?\\sin^2\alpha +cos^2\alpha =1\\cos^2\alpha =1-sin^2\alpha =1-(\frac{3}{4})^2=1-\frac{9}{16} =\frac{16-9}{16} =\frac{7}{16}.\\ cos\alpha =б\sqrt{\frac{7}{16} } =б\frac{\sqrt{7} }{4} .[/tex]

Ответ: cosα=±√7/4.