С подробным объяснением, пожалуйста
cos(x/2)=0
4+3cos2x=1
cos(6+3x)=-([tex] \frac{ \sqrt{2}}{2} [/tex])
cos(x/2)=0
4+3cos2x=1
cos(6+3x)=-([tex] \frac{ \sqrt{2}}{2} [/tex])
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Answers & Comments
x/2 = π/2 + πk, k∈Z
x = π + 2πk, k∈z
2) 4 + 3cos2x = 1
3cos2x = - 3
cos2x = - 1
2x = π + 2πn, n∈Z
x = π/2 + πn, n∈z
3) cos(6+3x) = - ()
6 + 3x = (+ -)arccos(- √2/2) + 2πk, k∈Z
6 + 3x = (+ -)(π - arccos√2/2) + 2πk, k∈z
6 + 3x = (+ -)*(π - π/4) + 2πk, k∈Z
6 + 3x = (+ -)(3π/4) + 2πk, k∈z
3x = (+ -)*(3π/4) - 6 + 2πk, k∈z
x = (+ -)*(π/4) - 2 + 2πk/3, k∈Z