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`        Please answer this trigonometry question correctly.`
5 months ago

```							hello student, this is an extremely ez ques.i am assuming the eqn is x^2 + 4 – 3cos(ax+b) – 2x= 0. write it as(x – 1)^2 + 3= 3cos(ax+b)here, max value of RHS is 3 whereas min value of LHS is 3 since (x – 1)^2 is always greater than equal to 0.so for equality to hold, (x – 1)^2 must be equal to zero which means that x=1.or 0+3= 3cos(a*1+b)= 3cos(a+b)or cos(a+b)=1= cos0this means that a+b= 0, 2pi, 4pi..... etc (as they cannot be less than zero).but both a and b are also less than equal to 3. so the max value of a+b is 6, which is less than 2pi.hence a+b can only be zero.so, the final answer would be zero.kindly approve :)
```
4 months ago
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