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How do you solve #sin(2x) - cos(x) - 2sin(x) + 1 = 0#?

Answer 1

Caroline Adams

#sin(2x) - cos(x) - 2sin(x) + 1 = 0#

#=>2sin(x)cos(x) - cos(x) - 2sin(x) + 1 = 0#

#=>cos(x)(2sin(x) - 1) -( 2sin(x) - 1) = 0#

#=>(2sin(x) - 1)( cos(x) - 1) = 0#

When #2sin(x)-1=0#

#=>sin(x)=1/2=sin(pi/6)#

#=>x=npi+(-)^npi/6" where "n inZZ#

Again when

#cos(x)=1#

#=>x=2kpi" where " k inZZ#

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Answer 2

Grace Ashcraft

To solve the equation sin(2x) - cos(x) - 2sin(x) + 1 = 0, you can use trigonometric identities and algebraic manipulation techniques. The equation can be simplified and rewritten as a quadratic equation in terms of sin(x). After finding the solutions for sin(x), you can then determine the corresponding values of x by using inverse trigonometric functions if necessary.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.