The number of value of x in [0,2π] satisfying the equation |cosx−sinx|≥√2, is
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Solution
Given equation is |cosx−sinx|≥√2 Since |cosx−sinx|≤√1+1=√2, ∴ we must have |cosx−sinx|=√2 ⇒∣∣∣cos(x+π4)∣∣∣=1⇒cos(x+π4)=−1,1 ∴x+π4=0,2π,4π,6π....π,3π,5π,... ∴x=3π4,7π4