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Click here👆to get an answer to your question ️ How do you find all the solutions in the interval [0, 2pi): 2cos^2(2x)1 = 0 ?
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The solutions are x=pi/3,pi,(5pi)/3 Use this identity: cos2x=2cos^2x-1 Now, here's the actual problem: 2cos2x+2cosx=0 2(color(red)(cos2x))+2cosx=0 2(color(red)(2cos^2x
Find all real numbers in the interval [0,2pi) that satisfy the equation. 3sec^2xtanx=4tanx
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