Number of roots of the equation $$sinxcosx+2+tan2x+cot2x$$ $$=3$$, x∈$$0$$,$$4$$π is

$$sinxcosx+tanx+cotx=3$$ $$sinxcosx+1sinxcosx=3$$ But $$sinxcosx+1sinxcosx$$ ≥$$2$$ Hence, no solution. ie; the number of real solutions is zero.

Trigonometric equations

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Question

$Number of roots of the equation |s i n x c o s x| + \sqrt{2 + t a n^{2} x + c o t^{2} x} = \sqrt{3}, x \in [0, 4 π] i s$

Easy

Solution

$|s i n x c o s x| + |t a n x + c o t x| = \sqrt{3} |s i n x c o s x| + \frac{1}{|s i n x c o s x|} = \sqrt{3} B u t |s i n x c o s x| + \frac{1}{|s i n x c o s x|} \geq 2 H e n c e, n o s o l u t i o n . i e; t h e n u m b e r o f r e a l s o l u t i o n s i s z e r o .$

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