7 sin^2+8cosx-8=0
помогите решить и выбрать корни [-pi/2;pi/2]
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7-7cos²x+8cosx-8=0cosx=a
7a²-8a+1=0
D=64-28=36
a1=(8-6)/14=1/7⇒cosx=1/7⇒x=+-arccos1/7+2πn,n∈z
a2=(8+6)/14=1⇒cosx=1⇒x=2πn,n∈z