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How do you solve? 5cos2x+3sinx=4

Answer 1

See the answer below....

#5cos2x+3sinx=4# #=>5(1-2sin^2x)+3sinx=4# #=>5-10sin^2x+3sinx=4# #=>10sin^2x-3sinx-1=0# #=>10sin^2x-5sinx+2sinx-1=0# #=>5sinx(2sinx-1)+1(2sinx-1)=0# #=>(5sinx+1)(2sinx-1)=0#

Now,either #" "sinx=-1/5" or. "sinx=1/2#

Now,you can determine the value of #x#.

Hope it helps.....thank you...

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Answer 2

Delilah Aguilar

To solve the equation (5\cos(2x) + 3\sin(x) = 4), follow these steps:

Rewrite the equation using trigonometric identities: (5(1 - 2\sin^2(x)) + 3\sin(x) = 4)
Distribute and rearrange the terms: (5 - 10\sin^2(x) + 3\sin(x) = 4)
Rearrange the equation to form a quadratic equation in terms of (\sin(x)): (10\sin^2(x) - 3\sin(x) + 1 = 0)
Solve the quadratic equation using factoring, the quadratic formula, or other suitable methods.

After solving the quadratic equation, you will find the values of (\sin(x)). Then, you can determine the corresponding values of (x) by using inverse trigonometric functions or considering the periodicity of trigonometric functions, depending on the range specified for (x).

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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.